Anish Chari makes a good point. Technically I should have put 0 as a digit in the multiplication table, or mention that 0 is a possibility for the single digit. The option of 0 does get excluded later when since there are multiple pairs with products ending in 0, and their sums are different. So 0 could not have been one of the digits.
I thought this was accounted for with the fact that the last digit of the product is the same as a house number, which cannot be zero as house numbers mist be positive integers (not including 0)
@@sewingmachine5467 its not that bryan doesn't remember his house number, its that WE don't know, but it doesn't matter because the sum will add up to 10 either way
@@sewingmachine5467 the video doesn't say that both numbers were his house number, but instead that there were 2 possibilities 21 and 9, obviously bryan would know whether its actually 21 or 9 because its his house, but whether his house number is 21 or 9, in either case the sum of the 2 numbers the girl is thinking of would add up to 10
Is 0 really a single digit number though? I mean 00 and 01 aren't two digit numbers. It seems to me that we define an N-digit number based on where the highest non-zero digit is...
I thought for sure you were going to point out, "Since Avery said 'the LAST DIGIT' of their product is the same as your house number", then the product must have more than one digit.
I thought the same. But that assumption makes the problem unsolvable (21,27 and 63 would all be possible if 3 and 7 don't count, and this gives different sums), I figured single digits must be allowed.
I got right to the end of the problem and decided that 3 + 7 was 12 and couldn't see how to decide if the pair was 1 and 9 or 3 and 7.... it is 4am but clearly I'm brain dead.
What gets me about this problem is the phrasing "the *last* digit of the product" - you don't say last unless there's more than 1 (well, outside of the saying "this is the first and last time I/you ____"). So that makes me think the product *HAS* to be two digits - but then that gives three possible number pairs: 3&7, 3&9, and 7&9. Since they don't add up to the same sum, that means the problem is unsolvable if you go by that logic.
If you're playing a "guess my numbers" game and the product was a single digit, would you just give that away? I think this would fit fine with the other exceptions.
@@badrunna-im I saw another comment (or a reply to another comment) that mentioned a better way to phrase it: "the product's *last OR only* digit". Gets rid of any confusion without revealing the answer.
@@Rainbowmon the last digit of a 1 digit number is just that 1 digit. If I ask you for a $20 bill, and you tell me "this is the last $20 bill I ever got from my workplace", I could make no inference about how many $20 bills you have, just that the one you're giving me is the last one you received.
I was a bit confused by the reference to a last digit. I thought that meant the product had to have two digits. But since that left the problem unsolvable (3 and 7 aren't eliminated as housenumbers), I got to the answer in the end.
I also thought the same, because no one would mention ‘last digit’ if the product is 9 (single digit). Then again they could have just asked for the number, not the sum of digits.
If you made a friend and he took you to walmart and then you pulled out a gun and robbed the place, and then the next day you asked him if he could take you back to walmart, and he said "No way, the last time I brought you to walmart you pulled out a gun and robbed the place!", would you then say "that's not true, you only took me to Walmart once"?
@@medexamtoolscom That's a sound argument on the surface. But it's false equivalency. It's a logic puzzle; "last digit" means there's more than one digit. Given that brings up an error when working out the problem, it has to be concluded the problem was worded incorrectly to start with.
For every pair of digits, there is another pair made up of their respective 10s complements. e.g. 2,7 and 8,3. The product of these pairs can be written as a•b and (10-a)•(10-b) respectively, which has the same last digit, the last digit of a•b. Any pair where the complementary pair is not itself leads to ambiguity, so they can be ruled out. What's left are the pairs that contain 10s complements themselves, so their sum would be, trivially, 10. Insight: 10 is the mean of 3 (1+2) and 17 (8+9), the minimum and maximum sums of two unique non-zero digits. Once you see the symmetry, the proof isn't too hard.
Wow, I was wondering if doing this puzzle in other bases would always lead to solution equal to the base, and I guess this means it would (assuming the puzzle _has_ a solution in that base, of course). Sweet.
I think I observed the same thing, only visually. The reverse diagonals on the chart are lines of constant sum. Each time he marked a group of numbers (having a common last digit) for elimination, you could see the pattern of placement of these numbers was symmetric along the reverse diagonal associated with the sum of 10. Any placement of numbers that included values off this diagonal would perforce have several different sums and so only a pattern contained entirely within that diagonal could be a solution, meaning the solution had to be 10. Turning this observation into algebra would give your result. Strangely, though, 3×7 was my first guess and turned out to be one of the correct ways of forming 10.
There's a shorter proof without touching the mean of 3 and 17. The product of the complements is 100 - 10(a+b) + ab, which, clearly, has the same final digit as ab (i.e., ab is equivalent, mod 10). Also simpler to handle with other bases than 10. Great insight to start with, though.
The numbers are 3 and 7. And the sum is 10. They are the only different numbers which product ends in 1. Edit: I didn't consider 1 and 9 were possible solutions too.
i thought that since the question says last digit, i assumed the product to be a 2 digit number so 3 and 7 makes sense , but anyways answer matters here that is 10. :)
Since Avery said “the last digit”, one would assume that they were referring to a double digit number, so 21. However, you’re right: they could have been mean and pedantic and meant 9, since the last digit of 9 is 9.
I was so close to solving this. I found that the last digits 1 and 9 only appear twice, which is okay because multiplication is commutative. So, yep, all I had to do was backtrack where the 1 and 9 came from and I would've had the answer. But instead, out of foolishness, I added them for some reason and it somehow happened to be 10 and I was correct out of luck. I found something by luck. The sum of the last digits of 21 and 9 is also 10.
The trouble here, is that the wording "The Last Digit" implies at least TWO digits. Which means you would eliminate all of the multiples of 1. That means that 21, 63, and 27, house numbers 1,3, and 7 respectively, are all unique. Without knowing which one is his, we get three possible answers: 10, 12, and 16.
@@mindtwist4738 I will admit that your math skills are exemplary, but your English might need some work. Last in a sequence of one is also the first but since first and last are contradictory, like any other double negative, they are omitted. Give me three examples in English where a single object of no sequence is referred to as the last object. If you saw a single person at the bus stop, you wouldn't ask them if they were the 'last one there' since you saw no one else. If you did, they would likely reply 'Only one' unless there had been others who had gotten onto other busses... but then, that would have been a sequence.
@@llywyllngryffyn8053 Let’s say you are hiring for a job opening and choosing the first qualified candidate. Now, if there is only 1 qualified candidate would you choose to hire no one since they aren’t the first? Similarly, if the last caller on a radio show receives a prize and there is only one caller, would you hand out no prizes since there is no last caller? I’m sure there are other examples as well. When dealing with a sequence of unknown length, randomly excluding certain vocabulary if the sequence was shorter than anticipated is silly.
My line of thinking was thus: The digits must both be odd, since having either or both of them as even would mean the product must be even. If the product is even, there are too many possible combinations for Bryan to be certain which two digits were used (eg a final digit of 2 could indicate a product of 12, which could be reached by 2x6 or 3x4, or a product of 32 (4x8). A final digit of 4 could be reached by 6x4 or 3x8, 6 could be reached by 2x3 or 2x8, 8 could be 2x4 or 2x9, and 0 could be 0xAnything, or any even multiple of 5). Thus both digits must be odd. Similarly, we can rule out the number 5 because any odd multiple of 5 will be 5, so there is no way for Bryan to determine what the other multiplier could be. That leaves us with the possibilities of 1, 3, 7 and 9. This works out to be possible products of: 1x3 = 3 1x7 = 7 1x9 = 9 3x7 = 21 3x9 = 27 7*9 = 63 Removing products that have the same final digit, this leaves: 1x9 = 9 3x7 = 21 Since Avery specifies the *last* digit of the product, logically the answer must be 3x7, and Bryan lives at house number 1.
@@oneoranota Or he got to the answer without using a multiplication table, but rather did it in his head with the correct thought process. Like I did. He is the one who decides whether to be happy about his result or not. Saying it was luck is possibly missing crucial details, meaning your reasoning is wrong.
The *last* digit of the product, would imply that the product is more than 10, which would exclude the number 9 - otherwise the wording should be "the digit of the product is the same as your house number"
Imagine Brian's house number was 7. Imagine the answer was 9+3 = 12 (9*3=27). Then Avery could have easily given this hint to Brian, as Brian would then deduce that "Last digit" should mean a 2-digit number. We, as external observers, cannot deduce what interpretation they would have followed considering the "last" digit. This could mean the answer could be 10 (is in the video) or 12 (9+3=12; 9*3=27) or 16 (9+7=16; 9*7=63) as they all have unique last digits in the multiplication table.
When Avery said, "The last digit of their product is the same as your house number," it's reasonable to infer that the product has more than one digit. For example if Avery's numbers were 9 and 1, would he really say "the LAST digit of the product is the same as your house number"? With that in mind we have three possibilities: There's only one way to get a product ending in 3 (9x7, with a sum of 16), and only one way to get a product ending in 7 (9x3, with a sum of 12), and only one way to get a product ending in 1 (7x3, with a sum of 10). Knowing his own house number, Bryan can solve the puzzle, but we cannot know which of the three possibilities to choose.
This was exactly the solution I got to. Also, an additional assumption I made after eliminating all single digit products (correctly or not) is that I eliminated all numbers with 0 as the second digit, since 0 is not used as a house number almost anywhere (of course there could be a extremely rare instance somewhere). In any case, those numbers get eliminated later since they’re not unique.
@@volodymyrgandzhuk361 Since the definition of last is: "coming after all others in time or order; final" and there are no other numbers. 9 is not the last digit of 9.
@@jorgehuertas3995 it's not? What is it then? Or maybe 9 has no last digit? Even your definition says "coming after ALL OTHERS", in plural, and by this logic, there have to be at least 3 digits
Even as a self-professed armchair logician, it seems I still have a long way to go in mastering the art of logic. I spent a good few minutes trying to work out the the minimum and maximum sums possible before realising I was supposed to be working out the products.
I applied fact that multiple pairs have the same sum earlier, which made solving this significantly easier to do and understand. The sums are constant over the diagonals (bottom left to top right, so 5+4=6+3), and the last digits are symmetrical with regard to the diagonal going through 9*1. So the solution must be on the symmetry axis and we don't even need to consider which house numbers are possible.
I don't know if it's been mentioned yet in the comments, but I'd assume that the phrase "last digit of their product" implies that the correct numbers would have a product greater than or equal to 10. This would make 3 and 7 the more correct answer of the two choices even if they have the same sum as 1 and 9.
But if you remove 2-to-9 from the possible solutions, there are other possibilities than 21 that are no longer eliminated by single-digit answers. Like 63, which only gets eliminated thanks to the presence of 3.
Then you would assume wrong. If I gave you one penny with the date "1986" on it and then asked you what the date on the last penny I gave you, would you say "your question makes no sense, because saying the last penny you gave me implies you gave me more than 1 penny!" No it doesn't. Saying the last of something doesn't mean there was ever more than 1 of them.
@@medexamtoolscom In that scenario, I'd have a go at you for using such poor grammar. :-) Asking what the date was on "the last penny I gave you" would indicate you had given me more than one penny, which would be false and make no sense. Correct grammar would have you using the definite article to correctly refer to the singular - "What is the date on THE penny I gave you?" As part of a logic puzzle, "last digit" means there is more than one digit.
@@mikee6666 Semantically though, "last digit" or "last penny" are still correct. You are *ASSUMING* there is more than one based on general human understanding and conventions, and that's a fair assumption of course, but just like when your parents would say "last warning" even though it was the first, and also the last, is also correct.
@@Fexghadi That was what I was explaining though - semantically the term "last penny" indicates there was a preceding penny for the "last penny" to follow, just as the term "last digit" indicates there is more than one digit. The assumption of there being more than one digit is based on the definition of the word "last". If whoever wrote out the puzzle did not intend for term "last digit" to mean more than one digit, then they made an incorrect choice of words. What you're saying is correct. If the term "last digit" is assumed to mean more than one digit, then the puzzle becomes unsolvable using the method demonstrated. Fortunately, as others in the comments section have shown, there are other ways of solving the problem that don't break the "last digit" logic.
I don't like this question, because "the last digit of the product is equal to your house number" carries the implication that the product has more than one digit (else Avery would simply say "the product is your house number"). Working under the assumption that the house number has two digits, hence ruling out all the single digits at the start, leaves 7x9=63 (sum 16) and 3x9=27 (sum 12) on the board, with no correct answer to the problem. Instead, Avery should say "the last or only digit of the product is equal to your house number"
This was my thought as well. Even saying "the ones digit is equal to your house number" would be enough to prevent ruling out single digit products. Given the language that was presented I automatically ruled out single digit products which left me with several possibilities but no definitive answer.
@@volodymyrgandzhuk361 First, yes that's exactly what it does. Second, the options for _3 are 3 and 63 (check the table yourself if you want). If you immediately discard all single digit numbers because of the misleading wording of the question, you are left with just 63 ending in 3. This is then a valid result to the problem (in addition to 21 and also 27) but the sum of digits is no longer the same
@@ospreytalon8318 no, it doesn't, even if literally, it seems like it should. It just means "the digit at the very right". But it doesn't automatically mean that there has to be another digit before it. Misleading wording? I have some news for you, many math problems are "misleading" by your definition, because the only other alternative would be to give away what one could find by reasoning. Would you also say that if there's a multiple-choice question and the last option is "none of the above", this implies that there actually exists an answer (different to all the given ones)?
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You have to draw the table to look at and check, it's too complicated if we don't have a pen and paper. My solution is that all digit numbers x (except 1) can be the production of 1 and x, so, if x is also the last digit of another production of two different numbers, we can eliminate x. For example, 2=1*2, but 2 also the last digit of 12 and 12=2*6, then we can eliminate 2. 3 is the last digit of 7*9=63, 4 is the last digit of 4*6=24 and so on. Check this rule for all 10 digits and we have only 1 and 9 are remaining. For 1, we have 7*3, for 9 we have 1*9 and 10 is the common sum.
Hi Presh, thanks for your postings, I'm a fan! I finally made solving this one but it took me some time. There's misleading information in the wording of this riddle. A says to B: "The last digit of the product is your house number". For me, this information does imply that the product of the two numbers MUST have two digits. If eliminating any product being only one digit there are three possible solutions: 3*7=21, last digit 1, 3*9=27, last digit 7, and 7*9=63, last digit 3. Each of the last digits are unique in the table, and each sum of the ones mentioned is different - so there would be 3 different solutions. After ignoring the thought the product HAS to be two digits and involving the one digit options I finally could solve it :)
Only problem with this is the phrasing of Avery’s last statement. That the _last_ digit exists implies there is more than one digit in the product, completely eliminating 1 from potentially being one of the numbers (and also 2,3 or 2,4 as solutions). Thus, 1, 3, or 7 could all be Bryan’s house number, which gives 3 options for solutions (3,7 or 3,9 or 7,9) and no way to exclude any of them.
Why? With a single digit result of multiplication the first digit and the last digit are just exactly the same digit - the only digit of the result. When the first person in a queue starts a queue of people are they in a queue or not? After all, there is, at this stage, only 1 person in the queue. Others join the queue behind the _last_ person in a queue, so if there is only 1 person waiting (the _first_ person in the queue) can they join the queue? If there is a sign "only the first person in the queue will be served" does that mean if only 1 person is waiting in the queue they will not be served until there is at least one other person in the queue?
@@cigmorfil4101 That's not comparable to the question's context. If Bryan lived at number 9, Avery has essentially said "the last digit of the product of my numbers is your house number" Why wouldn't Avery just say "the product of my numbers is your house number" in that case?
@@ospreytalon8318 But it is comparable to the understanding of "last". The reason being that this is a mathematic (sic) problem which has a unique solution which can be deduced from the given, sparse, information. Perhaps you would be happier to have final clue given as "the product modulus 10 is equal to your house number"? But how many people other than mathematicians would use that phraseology? Again "the units digit of the product..." is also a mouthful that is not really used in everyday speech. Perhaps we need to get a survey done of 11 year olds to ask them what they think the last digit of a number is. Just to point out, both those alternative versions are actually incorrect: the two different single digit numbers of which I'm thinking are .1 and .9 and multiplying them together gives .09 - there is nothing to say the two single digit *numbers* are *whole* numbers, and a leading zero (before the decimal point) isn't counted as its presence or not makes no difference to the value of the number, just like 09 and 9 both have the same value. And if the two single digit numbers are .1 and 9 the product is .9
Solve it by: 1. Since 1 x n = n, and the two numbers are different, then all digits except 1 must consist of at least one combination of the numbers. 2. Since 2 x n is an even number, then all even digits are eliminated. 3. Since the last digit of 5 x n is either 0 or 5, 5 is eliminated (0 is already eliminated). 4. Just check for the last digit of 3x7, 3x9, and 7x9, 3x7=21, 3x9=27, 7x9=63. So 3 and 7 are eliminated. 5. This gives the last digit is either 1 or 9, which consists of 3x7=21 or 1x9=9. Either of these results in the sum 10.
@@hans-rudigerdrzimmermann a sequence can have one, or even zero elements. All that having a 'last' element implies is that there is at least *one* element in it, but that's only because we assert that's there's an element in there for us to pick out.
@@mikee6666 that simply isn't true, if we're being as precise and mathematical as possible with our language. I accept that casually, speaking of a 'last' item usually suggests that there are at least three items, but mathematically, there need be only one to find a 'last' item, as I explained above.
I realize what Avery said is "the same number as your house". It means that you can have x² or ²√x if you multiply the same number on one diagonal side. And cross out the diagonal places or you can say crossing out the hypotenuse places because this table has a right angle and it is a scalene triangle that has a right angle. After all, you can multiply the same number more like repeated multiplication. So that is nice logic.
While working this out mentally, a helpful trick for me was to rule out all of the even numbers first (starting with zero) as Bryan's house number; then rule out 5, because these 6 options appear in so many products. This left only 4 choices and made it a lot easier to figure out without writing notes... which was helpful because I stink at keeping track of very many things at once without paper.
The answer to the question ,"Can you guess the sum?" is yes. I can ALWAYS guess the sum. Whether I can work it out is another matter. And whether I can guess correctly in a few attempts is another. Good maths, poor question.
I used the fact that ou writing system is base 10, so i eliminated all of the divisors(2, 5,1) and every even number(since if you multiplied it by 5, you would get a number ending in 0), this left 1,3,7 and 9, 3*9 and 7*1 AND 3*1 and 7*9 have the same last digit, so it was only left with 1*9 or 3*7, and their sum would be 10
Although he may have implicitly done so, I feel he needs to have also checked the sums of the numbers before eliminating them. For example, there may have been multiple pairs of numbers that had different products (all of which ended in 6, but were 16, 36, etc.) but their sums could have all been the same. This is easily checked (they would all have to lie on the same SW/NE diagonal, as 21 and 9 do), but it seems like he missed as step in his elimination process.
Yeah he should have mentioned this aspect, I think you should be able to show it cannot be if the number is 5 or smaller, since the smaller product of two different integers is 2*3. There could be a more rigorous way to show that it cannot happen for even numbers in general, but they all show at least twice in the same column meaning that the related sums cannot be equal in all cases.
@@pumaconcolor2855 although... the smallest product of two different integers is 0*1. And Presh also neglected to include the case where one of the 1-digit number was 0; even if that would get eliminated just like the other 7 digits with multiple products in the table, it should've been done..
When he says "the last digit...", doesn't it mean that there's more than 1 digits? So the product is a two digit number which gives us only 1 solution. Thoughts anyone?
In real life, that would be a good conclusion to draw from their word choice. In a math problem, it doesn't matter. A single digit still technically has a first and last digit (even though the first is the last) and these word problems will use that kind of wording/technicality to be intentionally vague.
If you make the assumption that the reference to the *last* digit of his house number means that it must be at least 2 digits long, then their product must be 21 (product of two primes), hence they are 3 and 7 and their sum is 10.
Ah, I had interpreted "last digit" as implying that the product had more than one digit, hence I could eliminate all products that resulted in only a single digit. This left both 3*9 (product 27, sum 12) and 7*9 (product 63, sum 16) as being potentially solvable by Bryan, but had eliminated 1*9 since it did not produce a 2-digit number.
In the multiplication table there is also a last digit symmetry on each side of the lower left to upper right diagonal (a = 10 - b). Any last digit besides the diagonal will appear at least twice. So all valid answers (24, 21, 16, 9) are on that a = 10 - b diagonal, we only need to check if their last digit appears on one side of the diagonal. 4 and 6 do, 1 and 9 do not. ==> Proof for the last digit symmetry around the a = 10 - b diagonal: Each cell (a, b) where a10 - b has a mirror cell (10 - b, 10 - a). Their products are ab and (10 - b)(10 - a). The difference ab - (10 - b)(10 - a) => ab - (10² - 10a - 10 b + ab) => ab - 100 + 10a + 10 b - ab => 10a + 10b - 100. As a and b are whole numbers the difference is always a multiple of 10 so the last digit is the same. That both sums had the same value is no coincidence either. On the diagonal a = 10 - b the sum of a + b is the same as the sum of (10 - b) and b which simplifies to just 10. I really like this problem as it revealed special properties of multiplication tables that greatly reduce the number of results to check.
In the problem, you mention that the last digit of the product of the numbers is the same as your house number. I interpret that to mean that that product had two digits -- not just one. I also ruled out the idea that his house number would have been '0' (Who lives in house 0?) This eliminates all single digit products as potential products. Using that logic, we are left with (3,7) to make 21 [house number 1] or a sum of 10, (3,9) to make 27 [house number 7] or a sum of 12, and (7,9) to make 63 [house number 3] with a sum of 16.
I was thinking like this -->Bryan is sure about the fact that the last digit is unique in the sense that it does not appear twice as the product of distinct numbers like 8.7=56 and 4.9=36 -->since both the digits are distinct so our range is from 0 to 72(9.8). --->Now numbers ending with the digit 1 are {1,11,21,31,41,51,61,71} 1 can be discarded using the fact that both the numbers are distinct and 31,41,61,71 can be discarded as the numbers are primes and 51=3.17 so it can also be discarded leaving us with the only possibility of 21=3*7. --->Now numbers ending with digit 9 are {09,19,29,39,49,59,69} out of these 19,29,59 are primes 39=3*13 49=7*7 leaving us with the only possibility of 09(1.9) and numbers ending with other digits can be discarded easily using counter examples as follows 0 can be discarded {10,20} 2 can be discarded {02,12} 3 can be discarded {03,63} 4 can be discarded {04,24} 5 can be discarded {05,15} 6 can be discarded {06,16} 7 can be discarded {07,27} 8 can be discarded {08,28}.
I would say this puzzle has no valid solution. There is nothing saying that the two numbers are non-negative. So every last digit appears in at least two products, a * b and a * -b. But that still gives no solution since a + b = a - b ⇔ b = 0 which implies that a * b = 0 but a house number can't be zero (and also zero appears as the last digit in more than one product).
@@shefali6599 But that is what I mean he can't do. If his house number was either 1 or 9, the sum could be -10, -8, -4, 4, 8 or 10. In general, for any house number he lives on, there are at least two different sums.
3:30 I don't agree that 7 can be eliminated, because Avery specifically says "the last digit of their product" implying their product is a multi-digit number. That means that if Brian lived on number 7, and interpreted "last digit" to imply a 2-digit number, then we can't say whether the result is 21 (i.e. 7+3=10) or 27 (i.e. 9+3=12). My conclusion is that this problem cannot be solved with the information given.
The sum is 10 because the only way he could guess the numbers was if his house number is 1 or 9. If the house number was 1 that would mean the numbers are 3 and 7. If the house number is 9 then the numbers were 1 and 9. Either way the sum is 10. If the house number is anything else there would be no way for him to determine the numbers. It's an easy puzzle, just a bit laborious to list out the possibilities. Fortunately the 2 numbers are interchangeable since multiplication and addition are commutative, otherwise it would be twice as much work.
As Avery had mentioned "last digit", if we consider the Product of the 2 numbers to have 2 digits? This would eliminate 1 as a possible number. Considering only the possibilities of 2-digit products, then the answer would be: 10 -> 3 & 7 if Bryan's house number is 1 OR 12 -> 3 & 9 if Bryan's house number is 7 OR 16 -> 7 & 9 if Bryan's house number is 3.
I eliminated the row of the 1's because no one says the last digit of your number when talking about a one-digit number. That left 21 as the only possibility of 3 and 7.
This problem make me remember a year ago, when I'm studying for this competition (I was grade 5 that day), and I found this problem. In summary, I did everything like you, but I have a faster way. First, we will try all possible house number, and find the possible product with this house number . If there are >1 numbers that can be the product of 2 different 1 digit number, eliminate it (Example: If the house number is 0, we see that 10 = 2*5, and 20 = 4*5, so we eliminate 0. If the house number is 1, the only possibility is 21 = 3*7, so 1 is a house number possibility) I think solve this problem by this way is faster, because it's will take a lot of time to calculate all the possibilities that is a product of 2 different 1 digit number. Thank you if you're reading this comment.
I disagree in part with this solution. A single digit product has no "last digit." Only two digit products can have a last digit. Therefore 3+7 is the only correct sim.
You are the first and the last in a group of one. I don't see why digits would be any different. A single digit product absolutely has a last digit unless you want to pretend that English words are more rigorous than math definitions...and English words don't really have a good track record for being that consistent, so that wouldn't be wise.
@@AThousandSunsinphysics in 09, 0 isn't taken as a significant digit. since its presence doesn't alter the original value. as a result you cant use that reasoning here
The very first step is to understand from their conversation , is that the product's last digit is UNIQUE, so by just knowing the last digit, one can know what's the solution. Then, note the number 1. It multiplies with all other numbers, and produces as results those numbers themselves(from 2-9). So either 1 is part of the solution, with as yet unknown number, OR if 1 is not part of solution then the result of product can only be 1 (which isn't produced by 1 above) I took the latter option first and found the solution... got lucky a bit.
Working out before I watch the video: 3-17, are the possible starting sums. 0 and 5 can be eliminated as house numbers since the second number can't be figured out with all the repeats. 1 seems like it wouldn't be one of the one digit numbers since A said Last digit.
One small problem, he said the "LAST digit" meaning the product consisted of more than one digit. So from the information we had his house number could have been 3 as there was only one product with two digits or more with a last digit of 3
In fact, the question is "can you guess their sum?" and the answer is "the sum of the two different one-digit number is 10". It doesn't matter the house number of Bryan or the fact that the sum of two different one-digit number has two possibilities, the right answer is 10. Thank you for sharing, great puzzle.
could also be negative 10.. we aren't sure at all (if you say last two digits can allow for singular digits then two numbers can be negative and validity of house numbers are irrelevant since context clues clearly are unreliable..)
A very fun puzzle that I was, coincidentally, able to solve by using my own house number, which ends in 1. Afterwards, I thought it was more of a "magic number" puzzle, so I tried old house numbers and none of them worked out.
The key is to find the last digit that occurs only once for all possible products. Since 1x(2..9) = 9x(8..1) mod 10, this eliminates 2 to 8 as last digit, leaving only 1 or 9 as possible last digit. So the product must be odd which means the original 2 digits are both odd and cannot be 5. That leaves only: 1, 3, 7, 9. There are only 2 possible solns: 1x9=9 or 3x7=21. And the sum in either case is 10.
Let x and y be two different integers in [1,9]. If x*y ends with the digit z, then (10-x)*(10-y) also end with the digit z. 10-x and 10-y are also two different integers in [1,9], and their sum is 20-(x+y), which is different from x+y iff x+y =/= 10. Thus, if x+y = 10, we can construct two different integers x' and y' in [1,9] such that x'y' has the same last digit than xy, and x+y =/= x'+y'. Thus, the only possibility is x+y = 10. This also shows that the solutions are on the "y=x axis".
There are two more things that, I believe, should have been considered - Home number that B is living in must be a 1 digit number, since "last digit" is equall to that number, therefore it is not 0, since there are no houses with Street number 0 (considering only average situation, no outliers or some individual made addresses). And the other information we got, that was not mantioned - the product is at least 2 digit number since "last digit of their product" - if A wasn't mean, the product is at least 2 digit. That way 1x9 is also out. Please correct me if my reasoning is wrong.
I had reasoned in a different way, and had a different result as a consequence. "the last digit" implies that there is more than one digit in the multiplication. Hence all single digit multiplications are eliminated. The last digit of the multiplication cannot be 0, since 0 is not a house number. This then results in 3 numbers that have unique last digits: 3*7=21 sum=10 3*9=27 sum=12 7*9=63 sum=16 In all 3 cases Bryan will know the sum of the numbers, since he knows which one of these cases matches his house number
The presented "solution" has a crucial error: The case "1 * 3 = 3" is out of consideration. ALL one-digit numbers are OUT OF CONSIDERATION! Since Avery in her second sentence gave away the information that the resulting number MUST HAVE 2 digits. Presh does not strike out the one-digit numbers as a preconditioning step, but should have done so! Doing that results in a undecidable situation: The "3" and the "7" are equally applicable as the last digit of the product: 27 and 63 are both equal cadidates. There is no information to make a choice between the two. Remember: A "solution" that builds on WRONG assumptions is NOT a solution.
The number nine has an infinite many preceding zeros (as do all numbers - they represent the zero-quantity of successive powers of ten in the number above the most significant digit). By convention we do not write those zero-digits, but they are there. Therefore, it is valid to say 'the last digit of 9 is 9'. Thus, without knowing Bryan's house number we can still arrive at a valid sum solution. It matters not what the house number is, only that it satisfies one of the two solutions.
So this generalizes to other bases (although not all bases will have a unique product). If you have two distinct numbers between 0 and one-less-than-the-base ("one-digit numbers"), then the only products that will have a unique last digit are of two numbers that sum to the base. That's because of how negative numbers work in modulus arithmetic. Example, 3*6=-3*-6, but -3*-6=7*4 (mod 10), so 3*6=7*4 (mod 10), meaning they share the last digit in base 10. If the two numbers sum to the base, then one is the negative of the other, so the above trick just gets you the same two numbers in a different order.
The puzzle says that the "last" digit of the product is the house number, which implies the product has two digits. 1 times any other one digit number does not equal a two digit number. EDIT--I just noticed that others have raised this point. I agree with those who point out that many logic puzzles depend upon very precise reading of the terms, and this one suffers from imprecision since the dictionary definition of "last" is "coming after all others in time or order"
There must be only one possible product with that particular last digit. So 0, 2, 4 and 8 can be ruled out as last digits, because both 1 and 6 give the same result when multiplied by these numbers. 5 and 6 can also be eliminated. Among 1, 3, 7 and 9, only 1 and 9 can remain, so it's either 3x7 or 1x9. Either way, the sum is 10.
I’m surprised I was able to figure this one out. I did an inverse approach to the multiplication table where I started by saying the house number is x, then seeing if there was more than one way to get a product with last digit x (I didn’t brute force this but just knew from experience/logic I could eliminate the 2-8). Leaving me with 21 and 9. I did mistakenly go the extra step by assuming the “last digit” statement meant it had to be 21, because 9 only has one digit and it would be strange to call that it’s last digit. However, this still lead to me to the correct sum 7+3=10
I think since the product is supposed to be a two digit number (Avery said "the last digit" implying there's a first one), we can eliminate every single digit product. Once we do it (and after following this video's method) we have two possible products: 21, meaning 3 and 7 are our numbers and therefore 10 the answer, and 27, meaning 3 and 9 are our numbers and 12 the answer. There's no way to know the actual answer so I think it's indeed an unsolvable problem. Please correct me if I'm wrong
Even assuming that Avery wouldn't have said "the last digit" for a one-digit number, I think you're wrong. Avery saying "the last digit" only implies that Avery's number has two digits, not that every possibility needs to have two digits. Bryan might not necessarily be making this assumption.
Haven't watched the video or read comments yet, but it looks like there are only two cases where the product ending uniquely describes a sum and both happen to have a sum of 10 (1 , 9) with digit 9 and (3, 7) with digit 1 so I'm saying 10 is the sum. His address would have to be either 9 or 1 for him to be sure of the answer.
The last digit of their product is the house number. Therefore the product should have two digits and therefore 3 and 7 will be the best-suited numbers.
I was able to figure this out on my own by some other method, by taking all the possible products(0*1,0*2.... 9*8,9*9)of all these numbers, I ended up with 32 values, of which I took the count of last digits in them and I got 9 as the only value which has occurred once and the guy was able to figure that out means he has got that unique value as his house number. Now the only factor for 9 is 1, so the sum is 9+1 =10. This approach is certain where you can get to know both the digits and the sum.
Lots of comments are saying that the product has to have two digits because the question mentions the last digit of the product, this is not true. The last digit of any number is the digit with no digits to its right, which for a single digit number is satisfied by the number itself. You could also say that the last digit is the digit such that all other digits come before it, and for a single digit number "all other digits" constitutes the empty set, so there is nothing to check and the statement is vacuously true.
Isn't 'sum' the result of an addition, so can't be a multiplication problem? That means the sum of two one digit numbers (1 to 9) would be between 2 and 18.
Hey, presh, technicaly you should have taken into account "the last digit" definition as a statement, that there MUST be at least TWO digits as result of multiplication! This is crucial for the solution you did.
Zero for sure should be eliminated but, no because of "many possibilities" but because of house numeration logic - normally we don't have "Zero" houses - these ones we should eliminate for sure - and that mean 0 digit in the table is absolutely redundant. Why by "The last digit of product is the same as your house number" in the decision it assuming that it should be unique/not repeating ("many possibilities" as said) in array ? I see no such condition. Bryan - has more data than we do - so he know his house number. E.g. if it's 5, so sum can be 8 (5+3) product 15 (5*3) - it's meats all conditions: 3 and 5 are one digit, they are different, and the Bryan knows his house number. I guess I am wrong but I tried to solve it more from practical point of view than mathematical.
Suppose the numbers are a and b. Then the product is ab. Consider the numbers (10-a) and (10-b). These will both be single digit numbers, so they could theoretically be the numbers as well. And since (10-a)(10-b) = 100 - 10a - 10b + ab = ab + 10(10-a-b), we know that the last digit is the same. So, the only way for this to work is if a = 10-a and b = 10-b, or a=10-b and b=10-a. If a=10-a, then a=5, and Bryan can't distinguish between 5*3 and 5*5, which give the sums 8 and 10. Also, b would also be 5 and we know the numbers are distinct. Therefore, a=10-b. Therefore, a+b=10. No need to brute force your way through.
I used an easier approach. The table of 9 (9, 18, 27…) hits all digits once. Since none divides 19, 29, 39, 49 (no need to higher, 8x7=56), we now know that “last digit 9” is unique. No need to check the rest or whether there are more combinations that fit to sum to 10.
The "last digit" can be a hint that the product is two digits and the answer can be 3 + 7, 3 + 9, 9 + 7. Knowing the house number, Bryan could easily solve the task.
@@igvarn I asked others and got no answer, I will now ask you: if there's a multiple-choice question and the last option says "none of the above", does it mean that there actually exists an answer (different to all the given ones)?
Doing these things, you learn to just figure out all possibilities, only one will actually work. So, I made a list of all possible options, figured out it was 1x9 or 3x7 and I had solved the puzzle before I even realized it's asking for a sum. House number is most likely 1, as it's kinda weird to say "last digit" when there only is one, but that's just bonus points.
last digit of product is house number could imply that product has 2 digits, removing all products < 10 ? With this rule, 3+7, 3+9 and 7+9 are all possibilities ?
I think this may have been presented in a bit of a misleading fashion. You made some inferences on Bryan's line of "Now I know!" As in, you took that to mean he knew one definitive answer and not just some quirky quote to wrap up the question. This was the only thing stumping me for a solid 5 minutes because I didn't think to look for a deeper meaning in that one liner. It's understandable that you used that as a piece of data, but perhaps the information could be put forth in a more explicit context. But with this in mind, knowing that we're to extrapolate some information rather than look at the surface level of the question, why not go further? Avery said that the last digit of the product of the numbers is equivalent to Bryan's house number. Since this dialogue is a conveyor for more information than could initially be assumed, Avery may have implied that the product she was thinking of had multiple digits when she said "The last digit." I'm aware that when properly defined, numbers can be seen as '...00xyz', but assuming Avery isn't a complete nerd and has a well formed vocabulary (based off of the technical and specific terms she used) it's a safe bet to say that she, perhaps even unconsciously, implied that there were multiple digits. This would reduce the amount of possibilities to just 1, giving an overall more easily solvable problem and a well more satisfying answer.
Anish Chari makes a good point. Technically I should have put 0 as a digit in the multiplication table, or mention that 0 is a possibility for the single digit. The option of 0 does get excluded later when since there are multiple pairs with products ending in 0, and their sums are different. So 0 could not have been one of the digits.
I thought this was accounted for with the fact that the last digit of the product is the same as a house number, which cannot be zero as house numbers mist be positive integers (not including 0)
@@sewingmachine5467 its not that bryan doesn't remember his house number, its that WE don't know, but it doesn't matter because the sum will add up to 10 either way
@@sewingmachine5467 the video doesn't say that both numbers were his house number, but instead that there were 2 possibilities 21 and 9, obviously bryan would know whether its actually 21 or 9 because its his house, but whether his house number is 21 or 9, in either case the sum of the 2 numbers the girl is thinking of would add up to 10
@@sewingmachine5467 he knows it(we don't) but whatever it is, the sum is 10
Is 0 really a single digit number though? I mean 00 and 01 aren't two digit numbers. It seems to me that we define an N-digit number based on where the highest non-zero digit is...
You got me. I couldn't get past "but Bryan knows his house number and I don't". The second you started eliminating cases I face-palmed.
I was able to solve it. I am very happy.
It was a hurdle to me as well, but I have the unfortunate habit of not stopping to solve these problems on my own most of the times.
I thought for sure you were going to point out, "Since Avery said 'the LAST DIGIT' of their product is the same as your house number", then the product must have more than one digit.
I had the same thought
That's not a valid argument. The last digit of 9 is 9.
@@Aiden-xn6wo but no one would specify the last digit when there's only one digit. It's not a thing anyone would say
I thought the same. But that assumption makes the problem unsolvable (21,27 and 63 would all be possible if 3 and 7 don't count, and this gives different sums), I figured single digits must be allowed.
The same! I'm very dissapointed for such inaccuracy.
No matter how advanced you get in mathematics, sometimes you just forget 1*7=7 for about 5 minutes.
Ikr
no?
😅 i also forgot about this but
Tried the problem again and
Solved the problem
I got right to the end of the problem and decided that 3 + 7 was 12 and couldn't see how to decide if the pair was 1 and 9 or 3 and 7.... it is 4am but clearly I'm brain dead.
My condolences, I solute you for doing math in the morning.
Presh: Here's a logic puzzle!
Also Presh: Let's brute force through all the possibilities!
Next logic puzzle: Name all 3-digit prime numbers!
That's a perfectly valid method for finite sets.
@@Tingletonttu would be nice if there was a more elegant solution
th-cam.com/video/LYr4RVf5uEw/w-d-xo.html
@@quixotix9540 It was elegant so....
Well, that went right over my head. Hats off to the 11 year old who got that question right.
Now, consider other bases.
Hehe
I am 11 years
@@kawsarjahan7282 joined when you were six?
@@ColinRichardson probably a parent's account
i love puzzles like this... where the fact that someone definitively figured it out is the key to figuring it out yourself
What gets me about this problem is the phrasing "the *last* digit of the product" - you don't say last unless there's more than 1 (well, outside of the saying "this is the first and last time I/you ____"). So that makes me think the product *HAS* to be two digits - but then that gives three possible number pairs: 3&7, 3&9, and 7&9. Since they don't add up to the same sum, that means the problem is unsolvable if you go by that logic.
I was going to answer the same thing
This was my answer as well
If you're playing a "guess my numbers" game and the product was a single digit, would you just give that away? I think this would fit fine with the other exceptions.
@@badrunna-im I saw another comment (or a reply to another comment) that mentioned a better way to phrase it: "the product's *last OR only* digit". Gets rid of any confusion without revealing the answer.
@@Rainbowmon the last digit of a 1 digit number is just that 1 digit. If I ask you for a $20 bill, and you tell me "this is the last $20 bill I ever got from my workplace", I could make no inference about how many $20 bills you have, just that the one you're giving me is the last one you received.
I was a bit confused by the reference to a last digit. I thought that meant the product had to have two digits. But since that left the problem unsolvable (3 and 7 aren't eliminated as housenumbers), I got to the answer in the end.
I also thought the same, because no one would mention ‘last digit’ if the product is 9 (single digit). Then again they could have just asked for the number, not the sum of digits.
100% agree (my answer was the same , for the same reason)
Same here. It was a bit deceptive but I got the right answer by including 1 and 9 as the last digit, since their factors both add up to 10.
If you made a friend and he took you to walmart and then you pulled out a gun and robbed the place, and then the next day you asked him if he could take you back to walmart, and he said "No way, the last time I brought you to walmart you pulled out a gun and robbed the place!", would you then say "that's not true, you only took me to Walmart once"?
@@medexamtoolscom That's a sound argument on the surface. But it's false equivalency. It's a logic puzzle; "last digit" means there's more than one digit. Given that brings up an error when working out the problem, it has to be concluded the problem was worded incorrectly to start with.
this is a great puzzle, love the reasoning involved, thanks for sharing
For every pair of digits, there is another pair made up of their respective 10s complements. e.g. 2,7 and 8,3. The product of these pairs can be written as a•b and (10-a)•(10-b) respectively, which has the same last digit, the last digit of a•b. Any pair where the complementary pair is not itself leads to ambiguity, so they can be ruled out. What's left are the pairs that contain 10s complements themselves, so their sum would be, trivially, 10.
Insight: 10 is the mean of 3 (1+2) and 17 (8+9), the minimum and maximum sums of two unique non-zero digits. Once you see the symmetry, the proof isn't too hard.
Wow, I was wondering if doing this puzzle in other bases would always lead to solution equal to the base, and I guess this means it would (assuming the puzzle _has_ a solution in that base, of course). Sweet.
I think I observed the same thing, only visually. The reverse diagonals on the chart are lines of constant sum. Each time he marked a group of numbers (having a common last digit) for elimination, you could see the pattern of placement of these numbers was symmetric along the reverse diagonal associated with the sum of 10. Any placement of numbers that included values off this diagonal would perforce have several different sums and so only a pattern contained entirely within that diagonal could be a solution, meaning the solution had to be 10.
Turning this observation into algebra would give your result.
Strangely, though, 3×7 was my first guess and turned out to be one of the correct ways of forming 10.
This is a wonderful solution, much more elegant than brute-forcing through all the possibilities!
There's a shorter proof without touching the mean of 3 and 17. The product of the complements is 100 - 10(a+b) + ab, which, clearly, has the same final digit as ab (i.e., ab is equivalent, mod 10). Also simpler to handle with other bases than 10.
Great insight to start with, though.
much better explanation than the video, really nice.
The numbers are 3 and 7. And the sum is 10. They are the only different numbers which product ends in 1.
Edit: I didn't consider 1 and 9 were possible solutions too.
Either way the sum is 10 (the answer that matters)
I actually mistakenly ignored the case "3 + 7", but found "1 + 9". Great minds think alike. 😀
Same lmao
i thought that since the question says last digit, i assumed the product to be a 2 digit number so 3 and 7 makes sense , but anyways answer matters here that is 10. :)
LOOOL you didn't consider 1 and 9 LOOOL
Since Avery said “the last digit”, one would assume that they were referring to a double digit number, so 21. However, you’re right: they could have been mean and pedantic and meant 9, since the last digit of 9 is 9.
My first thought was based on this reasoning
th-cam.com/video/LYr4RVf5uEw/w-d-xo.html
Huh, do you know any mathematically-inclined people like Avery and Bryan who *_aren't_* "mean and pedantic"? xD
Also, if the product had to be two digits, 9 would also be a valid answer (63). I will not find any more because I am too lazy.
I was so close to solving this. I found that the last digits 1 and 9 only appear twice, which is okay because multiplication is commutative. So, yep, all I had to do was backtrack where the 1 and 9 came from and I would've had the answer. But instead, out of foolishness, I added them for some reason and it somehow happened to be 10 and I was correct out of luck.
I found something by luck. The sum of the last digits of 21 and 9 is also 10.
The trouble here, is that the wording "The Last Digit" implies at least TWO digits. Which means you would eliminate all of the multiples of 1. That means that 21, 63, and 27, house numbers 1,3, and 7 respectively, are all unique. Without knowing which one is his, we get three possible answers: 10, 12, and 16.
"The Last Digit" does not imply that there are at least two digits. The last digit of a one-digit-number is simply the number itself
@@mindtwist4738 I will admit that your math skills are exemplary, but your English might need some work. Last in a sequence of one is also the first but since first and last are contradictory, like any other double negative, they are omitted. Give me three examples in English where a single object of no sequence is referred to as the last object. If you saw a single person at the bus stop, you wouldn't ask them if they were the 'last one there' since you saw no one else. If you did, they would likely reply 'Only one' unless there had been others who had gotten onto other busses... but then, that would have been a sequence.
@@llywyllngryffyn8053last and first aren't contradictory. A common phrase is: "that was the first and last time" (that I do X)
@@llywyllngryffyn8053 Let’s say you are hiring for a job opening and choosing the first qualified candidate. Now, if there is only 1 qualified candidate would you choose to hire no one since they aren’t the first? Similarly, if the last caller on a radio show receives a prize and there is only one caller, would you hand out no prizes since there is no last caller? I’m sure there are other examples as well. When dealing with a sequence of unknown length, randomly excluding certain vocabulary if the sequence was shorter than anticipated is silly.
My line of thinking was thus:
The digits must both be odd, since having either or both of them as even would mean the product must be even. If the product is even, there are too many possible combinations for Bryan to be certain which two digits were used (eg a final digit of 2 could indicate a product of 12, which could be reached by 2x6 or 3x4, or a product of 32 (4x8). A final digit of 4 could be reached by 6x4 or 3x8, 6 could be reached by 2x3 or 2x8, 8 could be 2x4 or 2x9, and 0 could be 0xAnything, or any even multiple of 5).
Thus both digits must be odd.
Similarly, we can rule out the number 5 because any odd multiple of 5 will be 5, so there is no way for Bryan to determine what the other multiplier could be.
That leaves us with the possibilities of 1, 3, 7 and 9.
This works out to be possible products of:
1x3 = 3
1x7 = 7
1x9 = 9
3x7 = 21
3x9 = 27
7*9 = 63
Removing products that have the same final digit, this leaves:
1x9 = 9
3x7 = 21
Since Avery specifies the *last* digit of the product, logically the answer must be 3x7, and Bryan lives at house number 1.
I realised 9 was the only product ending in 9 I didn’t consider 21 but I still got the correct answer of 10 so I’m happy
Nice I got exactly the opposite, didn't realise the 9, but thought of 21
@@simonvanprooijen because all the other prime numbers were still on the board, except 1
You should not be happy. Missing a crucial detail means your reasoning is wrong and you found the right answer by luck.
@@oneoranota Or he got to the answer without using a multiplication table, but rather did it in his head with the correct thought process. Like I did. He is the one who decides whether to be happy about his result or not. Saying it was luck is possibly missing crucial details, meaning your reasoning is wrong.
@@oneoranota cry
The *last* digit of the product, would imply that the product is more than 10, which would exclude the number 9 - otherwise the wording should be "the digit of the product is the same as your house number"
It doesn't imply anything. The last digit of 9 is 9. It's also its only digit, but being the only digit, it doesn't stop being the last digit.
Imagine Brian's house number was 7. Imagine the answer was 9+3 = 12 (9*3=27). Then Avery could have easily given this hint to Brian, as Brian would then deduce that "Last digit" should mean a 2-digit number. We, as external observers, cannot deduce what interpretation they would have followed considering the "last" digit. This could mean the answer could be 10 (is in the video) or 12 (9+3=12; 9*3=27) or 16 (9+7=16; 9*7=63) as they all have unique last digits in the multiplication table.
When Avery said, "The last digit of their product is the same as your house number," it's reasonable to infer that the product has more than one digit. For example if Avery's numbers were 9 and 1, would he really say "the LAST digit of the product is the same as your house number"?
With that in mind we have three possibilities: There's only one way to get a product ending in 3 (9x7, with a sum of 16), and only one way to get a product ending in 7 (9x3, with a sum of 12), and only one way to get a product ending in 1 (7x3, with a sum of 10). Knowing his own house number, Bryan can solve the puzzle, but we cannot know which of the three possibilities to choose.
This was exactly the solution I got to. Also, an additional assumption I made after eliminating all single digit products (correctly or not) is that I eliminated all numbers with 0 as the second digit, since 0 is not used as a house number almost anywhere (of course there could be a extremely rare instance somewhere). In any case, those numbers get eliminated later since they’re not unique.
Yes, he would. There's no reason to say that 9 is NOT the last digit of 9, or that 9 has NO last digit.
@@volodymyrgandzhuk361 Since the definition of last is: "coming after all others in time or order; final" and there are no other numbers. 9 is not the last digit of 9.
@@jorgehuertas3995 it's not? What is it then? Or maybe 9 has no last digit? Even your definition says "coming after ALL OTHERS", in plural, and by this logic, there have to be at least 3 digits
@@volodymyrgandzhuk361 Indeed, 9 has no last digit since it's a single item
Even as a self-professed armchair logician, it seems I still have a long way to go in mastering the art of logic. I spent a good few minutes trying to work out the the minimum and maximum sums possible before realising I was supposed to be working out the products.
I applied fact that multiple pairs have the same sum earlier, which made solving this significantly easier to do and understand.
The sums are constant over the diagonals (bottom left to top right, so 5+4=6+3), and the last digits are symmetrical with regard to the diagonal going through 9*1.
So the solution must be on the symmetry axis and we don't even need to consider which house numbers are possible.
I don't know if it's been mentioned yet in the comments, but I'd assume that the phrase "last digit of their product" implies that the correct numbers would have a product greater than or equal to 10. This would make 3 and 7 the more correct answer of the two choices even if they have the same sum as 1 and 9.
But if you remove 2-to-9 from the possible solutions, there are other possibilities than 21 that are no longer eliminated by single-digit answers. Like 63, which only gets eliminated thanks to the presence of 3.
Then you would assume wrong. If I gave you one penny with the date "1986" on it and then asked you what the date on the last penny I gave you, would you say "your question makes no sense, because saying the last penny you gave me implies you gave me more than 1 penny!" No it doesn't. Saying the last of something doesn't mean there was ever more than 1 of them.
@@medexamtoolscom In that scenario, I'd have a go at you for using such poor grammar. :-) Asking what the date was on "the last penny I gave you" would indicate you had given me more than one penny, which would be false and make no sense. Correct grammar would have you using the definite article to correctly refer to the singular - "What is the date on THE penny I gave you?"
As part of a logic puzzle, "last digit" means there is more than one digit.
@@mikee6666 Semantically though, "last digit" or "last penny" are still correct. You are *ASSUMING* there is more than one based on general human understanding and conventions, and that's a fair assumption of course, but just like when your parents would say "last warning" even though it was the first, and also the last, is also correct.
@@Fexghadi That was what I was explaining though - semantically the term "last penny" indicates there was a preceding penny for the "last penny" to follow, just as the term "last digit" indicates there is more than one digit. The assumption of there being more than one digit is based on the definition of the word "last". If whoever wrote out the puzzle did not intend for term "last digit" to mean more than one digit, then they made an incorrect choice of words.
What you're saying is correct. If the term "last digit" is assumed to mean more than one digit, then the puzzle becomes unsolvable using the method demonstrated. Fortunately, as others in the comments section have shown, there are other ways of solving the problem that don't break the "last digit" logic.
It's fun, few months ago I wouldn't be able to solve it but thanks those vids you make, I did it!
I don't like this question, because "the last digit of the product is equal to your house number" carries the implication that the product has more than one digit (else Avery would simply say "the product is your house number"). Working under the assumption that the house number has two digits, hence ruling out all the single digits at the start, leaves 7x9=63 (sum 16) and 3x9=27 (sum 12) on the board, with no correct answer to the problem.
Instead, Avery should say "the last or only digit of the product is equal to your house number"
This was my thought as well. Even saying "the ones digit is equal to your house number" would be enough to prevent ruling out single digit products. Given the language that was presented I automatically ruled out single digit products which left me with several possibilities but no definitive answer.
First, "the last digit" does not automatically imply there are several digits.
Second, the product can be 3·7=21, which has two digits.
@@volodymyrgandzhuk361 First, yes that's exactly what it does.
Second, the options for _3 are 3 and 63 (check the table yourself if you want). If you immediately discard all single digit numbers because of the misleading wording of the question, you are left with just 63 ending in 3. This is then a valid result to the problem (in addition to 21 and also 27) but the sum of digits is no longer the same
This was exactly my thought too. By this logic, we have other combinations like 3,9 and 7,9. Hence could not decipher the answer uniquely
@@ospreytalon8318 no, it doesn't, even if literally, it seems like it should. It just means "the digit at the very right". But it doesn't automatically mean that there has to be another digit before it. Misleading wording? I have some news for you, many math problems are "misleading" by your definition, because the only other alternative would be to give away what one could find by reasoning. Would you also say that if there's a multiple-choice question and the last option is "none of the above", this implies that there actually exists an answer (different to all the given ones)?
You have to draw the table to look at and check, it's too complicated if we don't have a pen and paper. My solution is that all digit numbers x (except 1) can be the production of 1 and x, so, if x is also the last digit of another production of two different numbers, we can eliminate x. For example, 2=1*2, but 2 also the last digit of 12 and 12=2*6, then we can eliminate 2. 3 is the last digit of 7*9=63, 4 is the last digit of 4*6=24 and so on. Check this rule for all 10 digits and we have only 1 and 9 are remaining. For 1, we have 7*3, for 9 we have 1*9 and 10 is the common sum.
I solved it without pen and paper. Did it all in my head. :D
Hi Presh, thanks for your postings, I'm a fan!
I finally made solving this one but it took me some time. There's misleading information in the wording of this riddle. A says to B: "The last digit of the product is your house number". For me, this information does imply that the product of the two numbers MUST have two digits. If eliminating any product being only one digit there are three possible solutions: 3*7=21, last digit 1, 3*9=27, last digit 7, and 7*9=63, last digit 3. Each of the last digits are unique in the table, and each sum of the ones mentioned is different - so there would be 3 different solutions. After ignoring the thought the product HAS to be two digits and involving the one digit options I finally could solve it :)
I remember this type of question when I was in grade 7. I was bad at the actual method so I just jammed 9 in everything.
Only problem with this is the phrasing of Avery’s last statement. That the _last_ digit exists implies there is more than one digit in the product, completely eliminating 1 from potentially being one of the numbers (and also 2,3 or 2,4 as solutions). Thus, 1, 3, or 7 could all be Bryan’s house number, which gives 3 options for solutions (3,7 or 3,9 or 7,9) and no way to exclude any of them.
Why?
With a single digit result of multiplication the first digit and the last digit are just exactly the same digit - the only digit of the result.
When the first person in a queue starts a queue of people are they in a queue or not? After all, there is, at this stage, only 1 person in the queue.
Others join the queue behind the _last_ person in a queue, so if there is only 1 person waiting (the _first_ person in the queue) can they join the queue?
If there is a sign "only the first person in the queue will be served" does that mean if only 1 person is waiting in the queue they will not be served until there is at least one other person in the queue?
@@cigmorfil4101 That's not comparable to the question's context.
If Bryan lived at number 9, Avery has essentially said "the last digit of the product of my numbers is your house number"
Why wouldn't Avery just say "the product of my numbers is your house number" in that case?
@@ospreytalon8318
But it is comparable to the understanding of "last".
The reason being that this is a mathematic (sic) problem which has a unique solution which can be deduced from the given, sparse, information.
Perhaps you would be happier to have final clue given as "the product modulus 10 is equal to your house number"?
But how many people other than mathematicians would use that phraseology? Again "the units digit of the product..." is also a mouthful that is not really used in everyday speech.
Perhaps we need to get a survey done of 11 year olds to ask them what they think the last digit of a number is.
Just to point out, both those alternative versions are actually incorrect: the two different single digit numbers of which I'm thinking are .1 and .9 and multiplying them together gives .09 - there is nothing to say the two single digit *numbers* are *whole* numbers, and a leading zero (before the decimal point) isn't counted as its presence or not makes no difference to the value of the number, just like 09 and 9 both have the same value. And if the two single digit numbers are .1 and 9 the product is .9
th-cam.com/video/LYr4RVf5uEw/w-d-xo.html
Solve it by:
1. Since 1 x n = n, and the two numbers are different, then all digits except 1 must consist of at least one combination of the numbers.
2. Since 2 x n is an even number, then all even digits are eliminated.
3. Since the last digit of 5 x n is either 0 or 5, 5 is eliminated (0 is already eliminated).
4. Just check for the last digit of 3x7, 3x9, and 7x9, 3x7=21, 3x9=27, 7x9=63. So 3 and 7 are eliminated.
5. This gives the last digit is either 1 or 9, which consists of 3x7=21 or 1x9=9. Either of these results in the sum 10.
I thought it was a little ambiguous because "the last digit of the product" implies at least a 2 digit number.
Why so? The last digit of a single digit number is just the number itself.
You are right , if the last digit is the same as the first digit it is a mistake. The last implies a series of numbers and not a single number.
@@hans-rudigerdrzimmermann a sequence can have one, or even zero elements.
All that having a 'last' element implies is that there is at least *one* element in it, but that's only because we assert that's there's an element in there for us to pick out.
Having "last" as a qualifier indicates pluralism; you can only have a "last digit" if there is a preceding digit.
@@mikee6666 that simply isn't true, if we're being as precise and mathematical as possible with our language.
I accept that casually, speaking of a 'last' item usually suggests that there are at least three items, but mathematically, there need be only one to find a 'last' item, as I explained above.
I realize what Avery said is "the same number as your house". It means that you can have x² or ²√x if you multiply the same number on one diagonal side. And cross out the diagonal places or you can say crossing out the hypotenuse places because this table has a right angle and it is a scalene triangle that has a right angle. After all, you can multiply the same number more like repeated multiplication. So that is nice logic.
"What if the last digit was equal to zero - that was Bryan's house number." Yes, yes - let's give this possibility the same consideration.
What if Bryan’s house number was 10, 20, or any other multi-digit number that ends in a 0
While working this out mentally, a helpful trick for me was to rule out all of the even numbers first (starting with zero) as Bryan's house number; then rule out 5, because these 6 options appear in so many products. This left only 4 choices and made it a lot easier to figure out without writing notes... which was helpful because I stink at keeping track of very many things at once without paper.
The answer to the question ,"Can you guess the sum?" is yes. I can ALWAYS guess the sum. Whether I can work it out is another matter. And whether I can guess correctly in a few attempts is another. Good maths, poor question.
In a "Pink Panther" movie, Inspector Clouseau asked a stranger if he knew what time it was.
The stranger said, "Yes."
The question is posed as a conversation between adolescents... Can u chill tf out
I used the fact that ou writing system is base 10, so i eliminated all of the divisors(2, 5,1) and every even number(since if you multiplied it by 5, you would get a number ending in 0), this left 1,3,7 and 9, 3*9 and 7*1 AND 3*1 and 7*9 have the same last digit, so it was only left with 1*9 or 3*7, and their sum would be 10
Although he may have implicitly done so, I feel he needs to have also checked the sums of the numbers before eliminating them. For example, there may have been multiple pairs of numbers that had different products (all of which ended in 6, but were 16, 36, etc.) but their sums could have all been the same. This is easily checked (they would all have to lie on the same SW/NE diagonal, as 21 and 9 do), but it seems like he missed as step in his elimination process.
Yeah he should have mentioned this aspect, I think you should be able to show it cannot be if the number is 5 or smaller, since the smaller product of two different integers is 2*3. There could be a more rigorous way to show that it cannot happen for even numbers in general, but they all show at least twice in the same column meaning that the related sums cannot be equal in all cases.
@@pumaconcolor2855 although... the smallest product of two different integers is 0*1. And Presh also neglected to include the case where one of the 1-digit number was 0; even if that would get eliminated just like the other 7 digits with multiple products in the table, it should've been done..
The couples of equal sums are on the lines parallel to the y=x axis. So you can visually tell if they all have equal sums or not.
Avery: I am thinking of a two different number, can you guess what are day.
Bryan: no I won't
Question ended
When he says "the last digit...", doesn't it mean that there's more than 1 digits? So the product is a two digit number which gives us only 1 solution. Thoughts anyone?
Fully agree. A failure on Presh's part, imnsho
It doesn't mean that at all. Last digit of number 6 is 6.
In real life, that would be a good conclusion to draw from their word choice. In a math problem, it doesn't matter. A single digit still technically has a first and last digit (even though the first is the last) and these word problems will use that kind of wording/technicality to be intentionally vague.
It is irrelevant. Queistion is what is the sum of numbers. It is 10 and does not matter whitch pair of numbers give this result.
@@jancermak1988 but it does matter which pair gives the result. The product of those two has "last" digit same as Bryan's house number.
If you make the assumption that the reference to the *last* digit of his house number means that it must be at least 2 digits long, then their product must be 21 (product of two primes), hence they are 3 and 7 and their sum is 10.
Ah, I had interpreted "last digit" as implying that the product had more than one digit, hence I could eliminate all products that resulted in only a single digit. This left both 3*9 (product 27, sum 12) and 7*9 (product 63, sum 16) as being potentially solvable by Bryan, but had eliminated 1*9 since it did not produce a 2-digit number.
Same...
Same, but also 7*3 (product 21, sum 10). (9*9) mod 10 is also 1, but because of the "different number" requirement, that wasn't in consideration.
In the multiplication table there is also a last digit symmetry on each side of the lower left to upper right diagonal (a = 10 - b).
Any last digit besides the diagonal will appear at least twice. So all valid answers (24, 21, 16, 9) are on that a = 10 - b diagonal, we only need to check if their last digit appears on one side of the diagonal. 4 and 6 do, 1 and 9 do not.
==> Proof for the last digit symmetry around the a = 10 - b diagonal:
Each cell (a, b) where a10 - b has a mirror cell (10 - b, 10 - a). Their products are ab and (10 - b)(10 - a).
The difference ab - (10 - b)(10 - a) => ab - (10² - 10a - 10 b + ab) => ab - 100 + 10a + 10 b - ab => 10a + 10b - 100.
As a and b are whole numbers the difference is always a multiple of 10 so the last digit is the same.
That both sums had the same value is no coincidence either.
On the diagonal a = 10 - b the sum of a + b is the same as the sum of (10 - b) and b which simplifies to just 10.
I really like this problem as it revealed special properties of multiplication tables that greatly reduce the number of results to check.
I miss Alice and Bob. I guess Avery and Brian does the job too :)
In the problem, you mention that the last digit of the product of the numbers is the same as your house number. I interpret that to mean that that product had two digits -- not just one. I also ruled out the idea that his house number would have been '0' (Who lives in house 0?)
This eliminates all single digit products as potential products. Using that logic, we are left with (3,7) to make 21 [house number 1] or a sum of 10, (3,9) to make 27 [house number 7] or a sum of 12, and (7,9) to make 63 [house number 3] with a sum of 16.
I failed this one because when I heard "last digit of their product", I assumed the product had to have at least two digits. 🙁
The coolest puzzle ever I've ever seen on this channel
I was thinking like this
-->Bryan is sure about the fact that the last digit is unique in the sense that it does not appear twice as the product of distinct numbers like 8.7=56 and 4.9=36
-->since both the digits are distinct so our range is from 0 to 72(9.8).
--->Now numbers ending with the digit 1 are {1,11,21,31,41,51,61,71}
1 can be discarded using the fact that both the numbers are distinct and 31,41,61,71 can be discarded as the numbers are primes and 51=3.17 so it can also be discarded leaving us with the only possibility of 21=3*7.
--->Now numbers ending with digit 9 are {09,19,29,39,49,59,69}
out of these 19,29,59 are primes
39=3*13
49=7*7
leaving us with the only possibility of 09(1.9)
and numbers ending with other digits can be discarded easily using counter examples as follows
0 can be discarded {10,20}
2 can be discarded {02,12}
3 can be discarded {03,63}
4 can be discarded {04,24}
5 can be discarded {05,15}
6 can be discarded {06,16}
7 can be discarded {07,27}
8 can be discarded {08,28}.
The first problem in this channel I've been able to completly solve!
I would say this puzzle has no valid solution. There is nothing saying that the two numbers are non-negative. So every last digit appears in at least two products, a * b and a * -b. But that still gives no solution since a + b = a - b ⇔ b = 0 which implies that a * b = 0 but a house number can't be zero (and also zero appears as the last digit in more than one product).
Key here is that Bryan in the end figured it out the sum, and that's where it can be known that it has a valid solution :)
@@shefali6599 But that is what I mean he can't do. If his house number was either 1 or 9, the sum could be -10, -8, -4, 4, 8 or 10. In general, for any house number he lives on, there are at least two different sums.
I saw the title to this video and immediately asked everyone in the room “does anyone know where bryan lives”
3:30 I don't agree that 7 can be eliminated, because Avery specifically says "the last digit of their product" implying their product is a multi-digit number. That means that if Brian lived on number 7, and interpreted "last digit" to imply a 2-digit number, then we can't say whether the result is 21 (i.e. 7+3=10) or 27 (i.e. 9+3=12). My conclusion is that this problem cannot be solved with the information given.
The sum is 10 because the only way he could guess the numbers was if his house number is 1 or 9. If the house number was 1 that would mean the numbers are 3 and 7. If the house number is 9 then the numbers were 1 and 9. Either way the sum is 10. If the house number is anything else there would be no way for him to determine the numbers. It's an easy puzzle, just a bit laborious to list out the possibilities. Fortunately the 2 numbers are interchangeable since multiplication and addition are commutative, otherwise it would be twice as much work.
Wow! Such a fantastic logical puzzle!!!
Einstein!
@@mustafizrahman2822 pam..
@@jimmykitty no paam. Truee always true.
@@mustafizrahman2822 Sobai tmr moto na..
@@jimmykitty Ha eita shotto bolso. Shobai amar moto posa satro na.
As Avery had mentioned "last digit", if we consider the Product of the 2 numbers to have 2 digits?
This would eliminate 1 as a possible number. Considering only the possibilities of 2-digit products, then the answer would be:
10 -> 3 & 7 if Bryan's house number is 1 OR
12 -> 3 & 9 if Bryan's house number is 7 OR
16 -> 7 & 9 if Bryan's house number is 3.
I eliminated the row of the 1's because no one says the last digit of your number when talking about a one-digit number. That left 21 as the only possibility of 3 and 7.
Or 63: 9&7
@@jordanbrock8856 good point
This problem make me remember a year ago, when I'm studying for this competition (I was grade 5 that day), and I found this problem. In summary, I did everything like you, but I have a faster way. First, we will try all possible house number, and find the possible product with this house number . If there are >1 numbers that can be the product of 2 different 1 digit number, eliminate it (Example: If the house number is 0, we see that 10 = 2*5, and 20 = 4*5, so we eliminate 0. If the house number is 1, the only possibility is 21 = 3*7, so 1 is a house number possibility) I think solve this problem by this way is faster, because it's will take a lot of time to calculate all the possibilities that is a product of 2 different 1 digit number.
Thank you if you're reading this comment.
I disagree in part with this solution. A single digit product has no "last digit." Only two digit products can have a last digit. Therefore 3+7 is the only correct sim.
At first I thought of it too. But then you can write 9 as 09 too. But yeah. 7 and 3 would be many people's choice..
A single digit number has a last digit of itself I disagree with you
You are the first and the last in a group of one. I don't see why digits would be any different. A single digit product absolutely has a last digit unless you want to pretend that English words are more rigorous than math definitions...and English words don't really have a good track record for being that consistent, so that wouldn't be wise.
@@AThousandSunsinphysics in 09, 0 isn't taken as a significant digit. since its presence doesn't alter the original value. as a result you cant use that reasoning here
If there is no house numbered with 0 then Avery was joking on Brian bc he's a bum
21 is the product (only product with 1 at the end), 10 is the sum for this (and 1x9, the only other product with 9 at the end)
As an 11 year old, im glad that i was able to solve it pretty quickly
The very first step is to understand from their conversation , is that the product's last digit is UNIQUE, so by just knowing the last digit, one can know what's the solution.
Then, note the number 1. It multiplies with all other numbers, and produces as results those numbers themselves(from 2-9).
So either 1 is part of the solution, with as yet unknown number, OR if 1 is not part of solution then the result of product can only be 1 (which isn't produced by 1 above)
I took the latter option first and found the solution... got lucky a bit.
I took Last digit to mean it was a 2 digit number, so I ruled out 1.
But either way the answer is the same!
Working out before I watch the video:
3-17, are the possible starting sums. 0 and 5 can be eliminated as house numbers since the second number can't be figured out with all the repeats.
1 seems like it wouldn't be one of the one digit numbers since A said Last digit.
Impressed by whoever came up with this puzzle!
One small problem, he said the "LAST digit" meaning the product consisted of more than one digit. So from the information we had his house number could have been 3 as there was only one product with two digits or more with a last digit of 3
In fact, the question is "can you guess their sum?" and the answer is "the sum of the two different one-digit number is 10".
It doesn't matter the house number of Bryan or the fact that the sum of two different one-digit number has two possibilities, the right answer is 10.
Thank you for sharing, great puzzle.
could also be negative 10.. we aren't sure at all (if you say last two digits can allow for singular digits then two numbers can be negative and validity of house numbers are irrelevant since context clues clearly are unreliable..)
A very fun puzzle that I was, coincidentally, able to solve by using my own house number, which ends in 1. Afterwards, I thought it was more of a "magic number" puzzle, so I tried old house numbers and none of them worked out.
The key is to find the last digit that occurs only once for all possible products. Since 1x(2..9) = 9x(8..1) mod 10, this eliminates 2 to 8 as last digit, leaving only 1 or 9 as possible last digit. So the product must be odd which means the original 2 digits are both odd and cannot be 5. That leaves only: 1, 3, 7, 9. There are only 2 possible solns: 1x9=9 or 3x7=21. And the sum in either case is 10.
Let x and y be two different integers in [1,9].
If x*y ends with the digit z, then (10-x)*(10-y) also end with the digit z.
10-x and 10-y are also two different integers in [1,9], and their sum is 20-(x+y), which is different from x+y iff x+y =/= 10.
Thus, if x+y = 10, we can construct two different integers x' and y' in [1,9] such that x'y' has the same last digit than xy, and x+y =/= x'+y'.
Thus, the only possibility is x+y = 10.
This also shows that the solutions are on the "y=x axis".
There are two more things that, I believe, should have been considered - Home number that B is living in must be a 1 digit number, since "last digit" is equall to that number, therefore it is not 0, since there are no houses with Street number 0 (considering only average situation, no outliers or some individual made addresses). And the other information we got, that was not mantioned - the product is at least 2 digit number since "last digit of their product" - if A wasn't mean, the product is at least 2 digit. That way 1x9 is also out. Please correct me if my reasoning is wrong.
I had reasoned in a different way, and had a different result as a consequence.
"the last digit" implies that there is more than one digit in the multiplication. Hence all single digit multiplications are eliminated.
The last digit of the multiplication cannot be 0, since 0 is not a house number.
This then results in 3 numbers that have unique last digits:
3*7=21 sum=10
3*9=27 sum=12
7*9=63 sum=16
In all 3 cases Bryan will know the sum of the numbers, since he knows which one of these cases matches his house number
The presented "solution" has a crucial error: The case "1 * 3 = 3" is out of consideration.
ALL one-digit numbers are OUT OF CONSIDERATION! Since Avery in her second sentence gave away the information that the resulting number MUST HAVE 2 digits.
Presh does not strike out the one-digit numbers as a preconditioning step, but should have done so!
Doing that results in a undecidable situation: The "3" and the "7" are equally applicable as the last digit of the product: 27 and 63 are both equal cadidates. There is no information to make a choice between the two.
Remember: A "solution" that builds on WRONG assumptions is NOT a solution.
Nowhere did Avery said that the resulting number must have 2 digits though.
th-cam.com/video/LYr4RVf5uEw/w-d-xo.html
The number nine has an infinite many preceding zeros (as do all numbers - they represent the zero-quantity of successive powers of ten in the number above the most significant digit). By convention we do not write those zero-digits, but they are there. Therefore, it is valid to say 'the last digit of 9 is 9'. Thus, without knowing Bryan's house number we can still arrive at a valid sum solution. It matters not what the house number is, only that it satisfies one of the two solutions.
So this generalizes to other bases (although not all bases will have a unique product). If you have two distinct numbers between 0 and one-less-than-the-base ("one-digit numbers"), then the only products that will have a unique last digit are of two numbers that sum to the base. That's because of how negative numbers work in modulus arithmetic.
Example, 3*6=-3*-6, but -3*-6=7*4 (mod 10), so 3*6=7*4 (mod 10), meaning they share the last digit in base 10.
If the two numbers sum to the base, then one is the negative of the other, so the above trick just gets you the same two numbers in a different order.
The instructions ask if the person solving can guess their sum. It doesn't state correctly guess. Also, what about -3, -7 solution?
This is the first one of your videos I’ve seen in which the solution was obvious to me
Avery: ''I am thinking of two different one-digit numbers, can you guess their sum?''
Me: Yes of course I can read your mind
The puzzle says that the "last" digit of the product is the house number, which implies the product has two digits. 1 times any other one digit number does not equal a two digit number. EDIT--I just noticed that others have raised this point. I agree with those who point out that many logic puzzles depend upon very precise reading of the terms, and this one suffers from imprecision since the dictionary definition of "last" is "coming after all others in time or order"
when the last digit of my house number is "A"
There must be only one possible product with that particular last digit. So 0, 2, 4 and 8 can be ruled out as last digits, because both 1 and 6 give the same result when multiplied by these numbers. 5 and 6 can also be eliminated. Among 1, 3, 7 and 9, only 1 and 9 can remain, so it's either 3x7 or 1x9. Either way, the sum is 10.
I’m surprised I was able to figure this one out. I did an inverse approach to the multiplication table where I started by saying the house number is x, then seeing if there was more than one way to get a product with last digit x (I didn’t brute force this but just knew from experience/logic I could eliminate the 2-8). Leaving me with 21 and 9. I did mistakenly go the extra step by assuming the “last digit” statement meant it had to be 21, because 9 only has one digit and it would be strange to call that it’s last digit. However, this still lead to me to the correct sum 7+3=10
I think since the product is supposed to be a two digit number (Avery said "the last digit" implying there's a first one), we can eliminate every single digit product. Once we do it (and after following this video's method) we have two possible products: 21, meaning 3 and 7 are our numbers and therefore 10 the answer, and 27, meaning 3 and 9 are our numbers and 12 the answer. There's no way to know the actual answer so I think it's indeed an unsolvable problem.
Please correct me if I'm wrong
Even assuming that Avery wouldn't have said "the last digit" for a one-digit number, I think you're wrong. Avery saying "the last digit" only implies that Avery's number has two digits, not that every possibility needs to have two digits. Bryan might not necessarily be making this assumption.
I got the same problem with different story in an Olympiad when I was 13.. Brutal !!
Haven't watched the video or read comments yet, but it looks like there are only two cases where the product ending uniquely describes a sum and both happen to have a sum of 10 (1 , 9) with digit 9 and (3, 7) with digit 1 so I'm saying 10 is the sum.
His address would have to be either 9 or 1 for him to be sure of the answer.
The last digit of their product is the house number. Therefore the product should have two digits and therefore 3 and 7 will be the best-suited numbers.
I was able to figure this out on my own by some other method, by taking all the possible products(0*1,0*2.... 9*8,9*9)of all these numbers, I ended up with 32 values, of which I took the count of last digits in them and I got 9 as the only value which has occurred once and the guy was able to figure that out means he has got that unique value as his house number. Now the only factor for 9 is 1, so the sum is 9+1 =10.
This approach is certain where you can get to know both the digits and the sum.
This is the first one of this guy's puzzles I've been able to solve on my own
Lots of comments are saying that the product has to have two digits because the question mentions the last digit of the product, this is not true. The last digit of any number is the digit with no digits to its right, which for a single digit number is satisfied by the number itself. You could also say that the last digit is the digit such that all other digits come before it, and for a single digit number "all other digits" constitutes the empty set, so there is nothing to check and the statement is vacuously true.
Isn't 'sum' the result of an addition, so can't be a multiplication problem? That means the sum of two one digit numbers (1 to 9) would be between 2 and 18.
Hey, presh, technicaly you should have taken into account "the last digit" definition as a statement, that there MUST be at least TWO digits as result of multiplication! This is crucial for the solution you did.
If it had to be at least two digits the then solution would be impossible.
Zero for sure should be eliminated but, no because of "many possibilities" but because of house numeration logic - normally we don't have "Zero" houses - these ones we should eliminate for sure - and that mean 0 digit in the table is absolutely redundant. Why by "The last digit of product is the same as your house number" in the decision it assuming that it should be unique/not repeating ("many possibilities" as said) in array ? I see no such condition. Bryan - has more data than we do - so he know his house number. E.g. if it's 5, so sum can be 8 (5+3) product 15 (5*3) - it's meats all conditions: 3 and 5 are one digit, they are different, and the Bryan knows his house number. I guess I am wrong but I tried to solve it more from practical point of view than mathematical.
Suppose the numbers are a and b. Then the product is ab. Consider the numbers (10-a) and (10-b). These will both be single digit numbers, so they could theoretically be the numbers as well. And since (10-a)(10-b) = 100 - 10a - 10b + ab = ab + 10(10-a-b), we know that the last digit is the same. So, the only way for this to work is if a = 10-a and b = 10-b, or a=10-b and b=10-a. If a=10-a, then a=5, and Bryan can't distinguish between 5*3 and 5*5, which give the sums 8 and 10. Also, b would also be 5 and we know the numbers are distinct. Therefore, a=10-b. Therefore, a+b=10.
No need to brute force your way through.
I used an easier approach. The table of 9 (9, 18, 27…) hits all digits once. Since none divides 19, 29, 39, 49 (no need to higher, 8x7=56), we now know that “last digit 9” is unique. No need to check the rest or whether there are more combinations that fit to sum to 10.
The "last digit" can be a hint that the product is two digits and the answer can be 3 + 7, 3 + 9, 9 + 7. Knowing the house number, Bryan could easily solve the task.
Last digit doesn't mean that there are two digits
@@volodymyrgandzhuk361 this means that there is more than one digit
@@igvarn no. It's just the remainder of the division by 10. The last digit of 9 is 9. Or are you saying that 9 has no last digit?
@@volodymyrgandzhuk361 if the number is from a single digit, then the "last digit" is a misleading hint
@@igvarn I asked others and got no answer, I will now ask you: if there's a multiple-choice question and the last option says "none of the above", does it mean that there actually exists an answer (different to all the given ones)?
I’m currently looking for a new job, so I’ve been watching a lot of these videos in hopes that one of these riddles is in the interview
I have figured out this puzzle on my own, but I have to say that it attributed to long hour viewing on your channels.😁
Doing these things, you learn to just figure out all possibilities, only one will actually work.
So, I made a list of all possible options, figured out it was 1x9 or 3x7 and I had solved the puzzle before I even realized it's asking for a sum.
House number is most likely 1, as it's kinda weird to say "last digit" when there only is one, but that's just bonus points.
The sum of the two numbers (10, in either case) is also the sum of his two possible house numbers (9 and 1).
last digit of product is house number could imply that product has 2 digits, removing all products < 10 ? With this rule, 3+7, 3+9 and 7+9 are all possibilities ?
I think this may have been presented in a bit of a misleading fashion. You made some inferences on Bryan's line of "Now I know!" As in, you took that to mean he knew one definitive answer and not just some quirky quote to wrap up the question. This was the only thing stumping me for a solid 5 minutes because I didn't think to look for a deeper meaning in that one liner. It's understandable that you used that as a piece of data, but perhaps the information could be put forth in a more explicit context.
But with this in mind, knowing that we're to extrapolate some information rather than look at the surface level of the question, why not go further?
Avery said that the last digit of the product of the numbers is equivalent to Bryan's house number. Since this dialogue is a conveyor for more information than could initially be assumed, Avery may have implied that the product she was thinking of had multiple digits when she said "The last digit." I'm aware that when properly defined, numbers can be seen as '...00xyz', but assuming Avery isn't a complete nerd and has a well formed vocabulary (based off of the technical and specific terms she used) it's a safe bet to say that she, perhaps even unconsciously, implied that there were multiple digits. This would reduce the amount of possibilities to just 1, giving an overall more easily solvable problem and a well more satisfying answer.