Birthday Dec 31. Conversation on Jan 1. Two days ago Dec 30 he was 13. One day ago turned 14. Today Jan 1 is a new year, will turn 15 at the end Dec 31 and be 16 at the end of the next year on the last day of the year.
I think it's cuz his birth (when he is 0 years old) is his 1st birthday and thus him turning 15 is his 16th birthday and fitting that into a coherent timeframe is trivial :)
I'd also say the same for the first one, but what they mean by "seemingly impossible" is that you might at first think it's impossible to be 13 two days ago and to turn 16 next year... The way that it sounds is odd, until you think about it. Perhaps to some people, the second one does sound similarly "seemingly impossible"
Puzzle 3 applies to me - I am almost exactly 27 years older than my son. When he was 03 years old, I was 30. Eleven years later, when he was 14, I was 41, and the pattern re-occurs every eleven years. I have thought a lot about this over the years, and it turns out that it will only work when the difference in ages is a multiple of 9, and it will always then repeat every eleven years.
Two days ago i was 13 (december 30) the very next day i turned 14 (december 31). Today is january 1 (it literally is) this year i'll turn 15 then next year ill turn 16. December 31 is my birthday. Happy new year!
@@kurisueru 13.5 years is thirteen and a half year. Some people might say 13 years 6 months. 13.99 is 14 years. But maybe you live on a different planet.
@@blackquintet Uh huh. And if you walk into a liquor store in the US saying you’re 20 years, 11 months, and 29 days old, the clerk will tell you to come back in a couple of days when you turn 21. Though I should have been more specific and said that age is expressed as an integer instead of measured. People that regularly walk around giving their age the way you describe are the ones often treated as though they’re from another planet.
On puzzle 3, I figured it out almost instantly because a few years prior I figured out that when you reverse a 2 digit number the difference is a multiple of 9
Whenever someone starts with "When I was your age..." again, I'm gonna throw them off with "Yes, I was your age less than twice my age back then." and leave them wondering.
@@peteneville698 Watch how he solved the 4th question again... If M is my age and S is someone's age, when that someone was M years old, i.e. S - M years ago... I would have been M - (S - M) = 2M - S years old back then.
Here's a simple solution I did for puzzle 4: Lets say that d is the age difference. Mary's age d years ago is half of John's current age and since the age difference itself is d, 2d is half of John's age, so d is 0.25 of John's age The sum of current ages is 1.75 Johns, and d years later, both ages will be 0.25 Johns higher, so it will be 2.25 Johns. Divide 63 by 2.25 to get 28, 28 * 0.75 = 21.
2:05 This 2nd puzzle is based on Married with Children, when Al Bundy was in a rare happy moment, and told his wife, "Next to the day Before I met you, This is the happiest day of my life!"
@@TruthTF Haha! I just saw a similar episode yesterday when Kelly got a job and Al told her, "This is the proudest moment of my life. And it shall remain so, until the day your mother leaves me."
Would any 14-year-old have the credentials (academic or otherwise) to be called a “logician”? Seems a stretch to use that title on someone that age. It would have made more sense to add, say, 30 years to each of the numbers in the problem.
NOTE: unless I say otherwise, I did not use the video to answer these questions. Puzzle 1 Birthday is on December 31st, and the riddle is on January 1st. 2 days ago, they were 13 (Dec 30) The next day, they turn 14 (Dec 31) The next day is new year, so this year they turn 15 and next year they turn 16. Puzzle 2 Yesterday = Wednesday Day before = Tuesday Day before = Monday 2 days before = Saturday 1 day after = Sunday Birthday was on Sunday. Puzzle 3 x = son’s age Age difference = x - 9 Assuming there is no illegal situation here, the dad should be at least 18 when the son was born. We can use trial and improvement to find the exact age. 18 + 9 = 27 (dad would be 72, and 27 and 72 are not 18 apart) Estimation: 35 - 9 = 26 (dad would be 53, and 35 and 53 are not 26 apart) Estimation: 36 - 9 = 27 (dad would be 63, and 36 and 63 ARE 27 apart.) Using maths and trial and improvement, we have figured out that the son is currently 36 and the dad is 63. Puzzle 4 John’s age = x Age difference between John & Mary = y Let’s rewrite the information so it makes more sense (or less sense if you don’t understand algebra lol) x = 2x - 4y 2x + y = 63 From the top equation, we can work out that x = 4y. This means that the bottom equation can be rewritten as: 8y + y = 63 Gather like terms: 9y = 63 Divide both sides by 9: y = 7 x = 4y = 4*7 = 28 28 - 7 = 21 John is 28, Mary is 21. Puzzle 5 I admit that for this one I watched part of the video, but only saw up to 12:07. You can infer that the census taker knows the house number, therefore there must be multiple ways to add up the children’s ages to get the house number and still get 36 when multiplying them. 36 as a product of prime factors is: 2*2*3*3. There are 6 possible age combinations for 3 children from this: 1, 1, 36 (sum 38, also unlikely as they are referred to as “children”, and a 36 yr old is an adult) 1, 4, 9 (sum 14) 1, 6, 6 (sum 13) 2, 2, 9 (sum 13) 2, 3, 6 (sum 11) 3, 3, 4 (sum 10) Out of these, only 2 share the same sum. Both combinations 1,6,6 and 2,2,9 have a sum of 13. However, as the mathematician refers to his eldest *child* (not children) and how he was excited to have *siblings* (not a sibling), we can infer that he is a good few years older than his siblings. As 1,6,6 would involve him having a twin, this combination is impossible, therefore, the eldest is 9 and the other 2 children are both 2 years old.
Riddle 5: Census Taker: Um, I have a long line of houses in which to take census after yours, and if they all give me riddles like this, I'll never get my job done.
It's quite nice seeing ukmt challenges, my secondary school offered them to me for 2 years in a row, and a boy who was in my class was the best in the whole south west!
I do the UKMT maths challenge every year at my school (only top set do it though) and last year I got through to the 2nd round (the only one in my year to do so). I didn’t get any further though
I did Puzzle 4 a bit differently. After working out John was 4/3 Mary's age, I realized that the sum was 9 "differences in age" - John is currently 4 "differences" old, Mary is 3 "differences" old, then you need to add one more for each of them the time Mary spends catching up to John's current age. 63 divided by 9 is 7, so that's the difference in their ages. This means John is currently four 7s old, making him 28, and Mary is currently three 7s old, making her 21.
I just divided 63 by 3, because I "know" Mary's current age has to be a third of the final sum. Then I double-checked that 14 was half of 28 and that 28+35=63. That "know" was a source of endless frustration for my teachers throughout the years. Certain things in word problems just seemed obvious to me, and whenever I tried to use formulae to show what I knew, I inevitably got the wrong answer.
The 5th puzzle is truly an all time classic, presented in different versions. There's a channel called Ted-Ed where they present riddles and the same riddle was presented in a life and death situation, where if you can't find a passcode, you'll be thrown into a cage of hungry salamanders😅
I'm glad you added the welcoming having siblings bit because just liking cookies really doesn't mean they have a unique age since there's always an elder among twins. But welcoming having siblings still isn't great, since it also means they'd need to be old enough to speak etc. when their siblings were born, as well as the fact that it can be read to imply they were welcoming multiple siblings at once (if I welcomed two siblings individually then at no point did I ever welcome siblings plural).
The eldest would not have been able to "welcome" siblings if they were a twin, since functionally they'd always be around one another. You can't be excited about gaining something you already possess.
you can welcome having siblings through gestures, not necessarily speaking, and you cant say in the pazzle "he is not old enough to talk" anyways so it doesnt matter. when someone says he welcomed having siblings doesn't mean they are twins, it's not a quote "thank you for my siblings", it could have been individually.
@@shakedeilstein2448 o for sure, I just meant it's still open to misinterpretation and feels a bit clunky. It's definitely an improvement on just saying the eldest likes something or other though!
The phrase in the riddle actually was “he was excited to have siblings” as in they were aware of not having siblings and looked forward to having some. Twins or other multiples would never be aware of not having siblings, as mydudestudios6244 stated. “Welcoming having siblings” would still refer to looking forward to having future siblings, not literally greeting them in any manner.
December 31st is the asker's birthday. The current day is January 1st, so two days ago was December 30th, the last day the asker was 13. The asker will be 14 for the entirety for the entirety for the current year, EXCEPT for the last day, December 31st, when they will be 15. They will then be 15 for all but the last day of the NEXT year, when they will turn 16.
You still never explain what “the eldest child loves chocolate chip cookies” has to do with solving the problem! If the mathematician had only said, “the eldest child was excited to have siblings”, without ever mentioning his son’s proclivity for cookies, the census taker would still have been able to derive the answer. From that statement alone, it is clear that the eldest child remembers a time in his life when he had no siblings-which would not have been the case if the kids were 6-6-1. Therefore, the only possibility is 9-2-2-regardless of whether any of the kids like cookies.
I think the important part there was the "My eldest child" refering to only 1 being the oldest and "he was excited to have siblings" referring to the youngest being twins (maybe)
No, saying you were excited to have siblings and remembering a time when you didn't doesn't tell you anything because that would have to always be the case. Of course there was a time when the oldest child was the only child, how could that ever not be the case?
verkuilb It's because the "excited to have siblings" part was added in later. To begin with, you were supposed to get the information that the oldest child isn't a twin from the phrase "eldest child" since most people don't think of one twin as being older than the other. Saying "loves chocolate chip cookies" was just an excuse to say SOMETHING about the oldest child. Then it was pointed out that not only could the oldest be a twin, but also he could be less than a year older than the next sibling, so they could be the same age without being twins. For example, my ex's two brothers were born in the same year, one in January and the other in December. So the detail about "excited to have siblings" was added in, indicating that the oldest would have to be over 1 year old when the next child was born. You're correct, this makes the chocolate chip cookies irrelevant, but whoever made up this version of the question didn't realize it and left it in.
Μπράβο ρε διάνοια πατριώτη!Εσύ πήρες όλο το 2ο Πεδίο κι'έφυγες για MIT!Εμείς τα "ταπεινά χαμομηλάκια" του 1ου Πεδίου,θα πάμε να κλάψουμε στην ορθή γωνία μας...
(Paused) I had a rough time parsing puzzle number 4 but I think I have the answer. John today is twice Mary's age when John was Mary's age. So if we say J is the age of John and M is the age of Mary, then what this line says is that when John was the age Mary is today, Mary's age would have been half the age he is now. So J = 2(M - (J-M)) which if you solve for J yields 4M/3 like so: J = 2(M - (J-M)) --> J = 2M - 2(J-M) --> J = 2M -2J + 2M --> J = 4M - 2J --> 3J = 4M --> J = 4M/3 When Mary will be John's age, the sum of their ages will be 63. To get Mary to John's age you would be adding the difference in their ages to her age, but John will also age the same amount in the same time frame, so you must add that difference to his age too. So 63 = (M + (J-M)) + (J + (J-M)). But we now know J is 4M/3 so I can just substitute that in for J and solve for M. 63 = (M + (J-M)) + (J + (J-M)) --> 63 = (M + ((4M/3)-M)) + ((4M/3) + ((4M/3)-M)) --> 63 = -M + 3(4M/3) --> 63 = -M + 4M --> 63 = 3M --> 21 = M So Mary is 21. And since we know J = 4M/3, then J=4(21)/3, or 28. So John is 28 and Mary is 21. When John was 21, Mary was 14 so he is twice that age now. When Mary is 28 John will be 35 and 28+35 does indeed = 63. Now to unpause and see if I got the right answer...
5:20 - We also know that the x > y (because the father's age 10x + y is greater than the son's age 10y + x), and we know that y is odd, because the father's age - 9 years is even (2 times the son's age - 9). We can also re-arrange 8x + 9 = 19y to 9 = 19y - 8x = 18y - 9x + y + x, which tells us that 9 = 9 (2y - x) + (x+y), which tells us that x+y must equal 9. All three constraints mean the father's age must be 81 or 63. From there it's trivial to see that 63 is the father's age.
@@jacobgoldman5780 1. The father's age is a 1-digit number. If the father is younger than 10, then the father is the same age as the son. This is an impossibility. 2. The father's age is a 3-digit number. The hundreds digit of the father's age must be greater than the ones digit, so the ones digit must be a 0; if the ones digit is a 1, the hundreds digit must at least be a 2, and the father would be 200+ years old. But the ones digit of the father's age must be odd, because the father's age 9 years ago was double the son's age 9 years ago. This is a contradiction: the ones digit of the father's age cannot be 0 *and* an odd number.
@@jacobgoldman5780 The only ages that work are: 1. The father and son are both 9 years old; 2. The father's age is 6 * 10^n + 3, the son's age is 3 * 10^n + 6. Examples: 63 & 36, 603 & 306, 6003 & 3006. The only pair that makes sense is 63 & 36.
I solved the last puzzle a bit theoretically: I wasn't really familiar with the concept of house numbers, so I skipped it (which kind of made my reasoning flawed). The eldest child had a distinct age, so it couldn't be 6,6,1. The child also loved choco-chip cookies, so it probably had well-grown teeth and hence it couldn't be 6,3,2; 4,3,3. The eldest was also excited to have siblings; so the other two could be twins. That left me with: 36,1,1; 9,2,2. Now 36,1,1 seems a bit far-fetched, so their ages are probably 9,2,2. The twin assumption is probably the greatest flaw, but the wording made me feel so.
If you misinterpret the odd framing of the question,it can imply they are the same age, so they would be 31.5 years old. Many people do this and I remember in my math club in elementary school this same problem became a point of argument.
You can read it as when John was Mary's age now, for example if John is 40 and Mary is 20 now then John is twice Mary's age now. Wording is not great but you can tell this is not the answer as the math doesn't work.
I keep interpreting it as, You are 13 and I am twice your age which makes me 26, when you are 26 I will be 39 and add them together it's 65 I know this isn't 63, but if you start with 12 the answer is 60 I'm still confused
Puzzle 1, you should pick the most extreme case possible to speculate. December 31 birthday, January 1st day of announcement. December 30th was 13. December 31, becomes 14. This year, 15. Next year 16.
so for puzzle 4: you are working with differences of ages, it makes total sense to declare some variable to use for it. Like d = J - M (as John is the older one) We know that M = J - d and J = 2(M - d), so J = 4d and M = 3d We also know that J (Mary's age when she is as old as John is today) + (J + d) (John's age at that time) = 63 so 2J + d = 63 -> 2(4d) + d = 63 -> 9d = 63 -> d = 7 which means J = 4*7 = 28 and M = 3d = 3*7 = 21
Puzzle 4: I think it's less confusing to keep the first equation as 3J=4M and to substitute for 3J instead of J in the equation 3J-M=63. That way you directly have 4M-M=63. It's pretty much the same thing but deviding in equations like this always confuses me. Great video btw (:
For the fourth riddle, I found it easier to put the past ages at J and M and the present ages as J+x and M+x, making the future ages J+2x and M+2x. In the end, I had converted everything to xs and just had to divide 63 by 9 (the number of xs) to get to x=7.
For the first riddle, although your answer is probably the correct solution. I read the riddle differently. I would argue you become 15 years old when you have your 16th birthday. Your first birthday is technically the day of your birth. 1 year later you turn 1 year old by having your 2nd birthday, it's just the 1st anniversary of your birth. Though you can get more technical about it and argue you become 15 years and 9-10 months old on your 16th birthday. Your answer is correct. I just think it's funny to think of it like this.
I think that 4th puzzle could definitely be worded better, specifically the first sentence, it should be "John today is twice the age that Mary was when John was Mary's age"
I spent a long time getting John and Mary wrong! Happy I got the others though, the census one is good but needs a bit of assumption that the visitor is a genius.
Thumbnail/First puzzle: December 31. If it is January 1, the girl was 13 on December 30. She turns 14 on December 31 and is still 14 on January 1. Next year, she'll be 16 because her birthday will have already passed again (making her 15) by the time the new year rolls aground. Before that year ends, she'll be 16. Second puzzle: Sunday. The day before the day before yesterday is three days before today. Today is Thursday, so three days prior is Monday. Monday was two days after the day before the birthday. Monday is two days after Saturday, so Saturday was the day before the birthday, thus the birthday is Sunday. This wasn't wasn't really a puzzle as much as it was just parsing through and counting the number of "day before"s. Third puzzle: The father is 63 and the son is 36. The father and son can only have so many ages, such as 91 and 19, 92 and 29, etc., but most will not fit the rule for the doubled age. Any triple-digit ages can be disregarded for this reason. Realistically, the easiest way of solving this is to brute-force it with Excel or Google Sheets, using formulas to fill it out. For example, A1 is 1, B1 is A1*2, C1 is A1+9, D1 is B1+9, and E1 is FLOOR(D1 / 10, 1) + (MOD(D1, 10) * 10) - C1. From there, autofill the spreadsheet. If E equals 0, there's the answer. Starting at 1, the answer will be on the 27th row. Fourth puzzle: John is 28 and Mary is 21. I'll get to the lengthy math after the fifth puzzle. Fifth puzzle: Their ages are 2, 2, and 9. I'm pretty sure you've done a video on this one before, but there were arguments in the comments about how "my oldest loves chocolate chip cookies" does not disqualify being a twin. There are only so many combinations of three numbers with a product of 36. Of them, their sums will produce a list of possible house numbers. The census taker knows the house number. If the correct house number appeared only once in the list, the census taker would know the answer. Since they did not, the house number must have appeared more than once. I broke out the spreadsheet again to help simplify finding this. The only sum that appears more than once is 13, using the ages (1, 6, 6) or (2, 2, 9). The cookie hint indicates there is an oldest child that does not have a twin, therefore it must be the latter option. With that, I got all five puzzles. Fourth puzzle explained: I do not like the wording on this one, but I think this is what it is saying: John is older than Mary. John's current age is an even number. When Mary age was half John's current age, John's age then is Mary's age now. When Mary reaches John's current age, their ages will have a sum of 63. Jf = John's future age for a sum of 63 Mf = Mary's future age for a sum of 63 J = John's current age M = Mary's current age Jp = John's past age Mp = Mary's past age 63 = Jf + Mf Mf = J M = Jp J = 2 * Mp Jf - J = Mf - M J - Jp = M - Mp J - Jp = M - Mp J - M = M - Mp J = 2 * M - Mp 2 * Mp = 2 * M - Mp 3 * Mp = 2 * M 1.5 * Mp = M 63 = Jf + Mf 63 - J = Jf - J + Mf 63 - J = Mf - M + Mf 63 - J = 2(Mf) - M 63 - J = 2(J) - M 63 - (2 * Mp) = 2(2 * Mp) - M 63 = 3(2 * Mp) - M 63 = 3(2 * Mp) - 1.5 * Mp 63 = Mp * (6 - 1.5) 63 = 4.5 * Mp 14 = Mp J = 2 * Mp J = 2 * 14 J = 28 M = Jp J - Jp = M - Mp 28 - M = M - 14 42 = 2 * M M = 21
I have never once heard the word distinct to describe an age. It's pretty much never used for that. Distinct means something stands out, is definite, recognizable, as if it's been highlighted somehow. I think the word you're looking for is unique.
I solved nos 4 and 5 differently. For 4, I realized that when Mary is John's age today the sum of their ages would be 63 and that means 3J-M=63. So, I just made a 3 column table and found with the columns "J", "M" (calculated as 3J-63), and "M2" (which was Mary's age when John was Mary's current age. calculated as 2M-J). And I just had to go down until I saw an answer where J is double M2. J - M - M2 31 - 30 - 29 30 - 27 - 24 29 - 24 - 19 28 - 21 - 14 And there was my answer. For 5 I'm not sure this was a valid solution, but I got the same answer as Presh. I realized that the answer had to be something where the oldest child had a different age than the other siblings, because they were excited to have siblings, which means there's a time they can remember where they didn't. I also figured the two younger had to be twins, because they were excited to have siblings (plural). So, I figured the only possibilities were 1, 1, 36; 2, 2, 9; and 3, 3, 4. I figured a 1 year old would be too young to comprehend the fact that their father's baby mama was pregnant and get excited. And I figured that "loves chocolate chip cookies" isn't the first thing you'd mention about a 36 year old. So, I got 2, 2, 9.
So, if you wanted to do it completely algebraically without using the table method you would realized J=2(2M-J), which means 3J=4M and 3/4J=M. Also, J+M=63 So, 7/4J=63. Therefore J=28.
For puzzle 3 I had to write a bit of code to check stuff before I found it, then I saw you doing the equations and thought "Oh there's gunna be some easy way to do this" and then you pull out the spreadsheet and check every option just like I did.
For the final puzzle, we're assuming the census taker is a mathematician as well. His response "I don't know" tells me nothing, as that's probably the answer of any census taker when given such an absurd answer.
The fact that the census taker says "Ah, now I know" after the mathematician started babbling about how the eldest loves chocolate chip cookies, indicates that the census taker is also a math aficionado.
@yurenchu Haha or he jotted down on his notepad "Crazy person loves here" and got out of there as soon as possible. Of course you're right, just being facetious.
For puzzle 3 I did it this way: F is the father’s age and S is the son’s age F - 9 = 2(S-9) Which simplifies to S = (F+9)/2 Then you can graph it and find a value where S and F are the same number only reversed. There are other ways to do it too, but this was what came to my mind first for some reason.
For puzzle 4, I had more variables, one for each age in each timeframe (past, present, future), and more systems of equation. I managed to rearange and substitute through to isolate John's age, then solve for Mary's age.
The first puzzle I interpreted differently. Based on the wording, the logician is 14 years old/14 years of age presently being she either is still 13 and mentioning she was also 13 years old two days ago, or she is 14 years old presently having a birthday over the past two days. She has had 15 birthdays once reaching 14 years of age (including the day she was born, 0 years old), so next year when she turns 15 years old it will also be her 16th birthday.
If the eldest child were 6 years 11 months old and the second-eldest 6 years 0 months old, the eldest child wouldn’t have learned to articulate such a complex thought as “I’m excited to have siblings” at less than a year old. That’s the new assumption
For the fifth puzzle, I don't believe knowing that there is an oldest child means knowing that the oldest and second-oldest children have different integer ages. I am an uncle to two sisters who were born ten months apart, so for two months every year, they have the same integer age as one another, but there is clearly one who is older and one who is younger. Even in the more common example of twins, one baby generally arrives before the other one, even if it's only by a matter of minutes. The only scenario I could imagine is if the twins are born through C-section, and somehow both babies were extracted at the same time. I don't know enough about that process to know if that is even possible, but if it's possible, then there would be no oldest child, but two oldest children.
I think the relevant bit is actually that "he was excited to have siblings." Siblings, plural, as in twins. If they were only expecting 1 child, then he'd only have been excited to have **a** sibling.
Saying he was excited to have siblings does not mean that they arrived at the same time. When he was an only child, he could have said, “I can’t wait to have brothers and sisters”, without knowing if/when they might arrive, or in what quantities. However, the fact that he was excited to welcome siblings DOES indicate that there was a time when h he was an only child, and was old enough to be aware of that fact.
It isnt. This has nothing to do with the puzzle. When you turn 1, it is your FIRST birthday. Birthday means birthday anniversary - not the literal day AND year you were born. Noone has ever celebrated a newborn babies birthday of the 0 years of life milestone.
Strictly, you are 21 on the day before what is commonly called your 21st birthday: so; if you were born on 21st June 2004 the 21st June 2025 is the 22nd time you have been alive on 21st June- you had been alive 20 years and 365 days immediately after whatever time you were born on 20th June.
Birthdays are the years weve finished. If you say "Im 21." that means youve finished 21 years, and started your 22nd. On the day of your birth is your 0th birthday bc youve finished 0 years and are starting the 1st.
The second puzzle is a lot easier for russian language speakers, because we have separate names for the "the day before yesterday" and "the day after tomorrow"
The first puzzle has a second solution (actually I would argue this is the _correct_ solution): your 1st birthday is the day you are born, when you are 0 years old. Thus, your Xth birthday is the day you turn X - 1 years old. Therefore your 16th birthday is the day you turn 15. It's the same as how the 20th century is the 1900s rather than the 2000s despite the latter being the one that actually starts with 20. So we'd have: 2 days ago, they were 13. Then they turned 14, but it was the 15th birthday they had. Next year they'll have their 16th birthday, when they will turn 15.
Also for 5; "...he was excited to have siblings" suggests that twins were arriving (rather than having "a sibling" or "another sibling") so the younger two would be the same age?
Well, since (half) my post got deleted, I'm just gonna paraphrase, because I'm lazy. Back in the '90s, when I was a sophomore in High School, my math teacher somehow divined that I was bored to death with my Algebra II classwork, and approached me privately with an almost exact copy of the Census Taker problem you laid out here (the man was in his 90s, to allow for the possibility of a 72-year-old child, and the "oldest child" was limited to liking chocolate cake; the "enjoyed having siblings" remark was simply unnecessary, as it probably is here). My immediate reaction to reading the problem was, 'I can't possibly solve this, there isn't enough information." But, my teacher said that there actually was, so I basically brute-forced the issue and had the correct answer within twenty minutes. Still, to this day, I love the way the problem is constructed, since the census-taker's ignorance is obviously important but comes across as being so pointless. A few years later, this very problem inspired me to construct my own "unsolvable" problem that is totally solvable and yet, unlike the Census Taker problem that only _seems_ impossible at first glace, actually seems completely unsolvable from any possible perspective, and I did it by including, as one of if not _the_ core aspect of the problem, a premise that seems on its face to be plainly and incontrovertibly just flat-out _wrong._ I came up with this problem three decades ago. I don't know how many people have read it, but I posted it back on Slashdot, TotalFark, SomethingAwful, the math and physics forums on Yahoo Answers, etc. Not only has no one, as of this very day of January 5th, 2025, ever gotten the correct answer, I don't think I've ever seen anyone employ the sort of approach that could even _lead_ to the correct answer. Now, there _is_ a correct answer, and despite giving zero hints when I first posted it, I _will_ save a _lot_ of time by letting you know that neither zero nor infinity are the answer, or even _appear_ in the answer. The answer is a positive, finite value, and that's all the clues you're getting. Now, since my original post screwed up, I'm gonna try posting this part first, and _then_ edit the actual question in more or less from unchanged from my original document. Oh, one more hint: this question employs geometry, but it is a purely _mathematical_ question. Physics are _involved_ in the question, but they're not _relevant_ to the question, nor are they to the answer. This is pure math. Here we go: -=-=-=-=-=-=-=-=-=-=-=-=-=- *The Flat Sphere* First, we must consider a hypothetical sphere of unspecified and irrelevant physical attributes. Only a few basic properties matter: 1) It is a physical object. 2) It exists somewhere in the universe. 3) It is solid, or atleast has a solid surface. 4) It is perfectly spherical. 5) Its size is dynamic (explanation below). So, a completely round solid spherical object... a cue ball from a pool table suffices perfectly for this hypothetical object. Let us also assume that physical forces such as gravity, electromagnetism, inertia, and angular momentum are not at work, or atleast that this object is immune to them and they are irrelevant for our purposes; our cue ball is not going to be collapsing under its own gravity or breaking apart due to shearing forces. Our sphere, for that matter, is not even comprised of atoms or other particles, but is constructed merely from a solid material. While it exists somewhere in our universe, let us consider it to be indestructible. And, finally, to the crucial element. We can take this sphere and increase or decrease its diameter, and thus its radius, circumference, surface area, volume, etc., at will. Although it really makes no difference, let's assume it begins at the size of a normal cue ball, a few inches in diameter. From here, we can manipulate it and alter its size down to the diameter of a neutrino or smaller, or up to the diameter of the Milky Way or larger. We can make it as large or as small as we possibly want. Now, here is the riddle: As the diameter of a sphere increases, its curvature decreases, and vice versa. So, what size does our object have to be in order for the surface of the sphere to become *completely flat*? There is, in fact, despite first appearances and protests to the contrary that I have received in the past, a correct answer, and only *one* correct answer, to this riddle. In addition, *all* of the information necessary to arrive at the correct answer has been provided here, so I’m afraid no further help or details can be given. Best of luck!
I remember a puzzle similar to puzzle 5; I don't know the numbers involved or the setup - but the end of the puzzle is something like Mathematician A "I don't know" followed by Mathematician B saying "Now I know" followed by Mathematician A saying "Now I know, too!"
he did a few of this "i dont know" "oh, now i know" in this channel, but he did an even similar problem in this channel, it was pretty much the same just with 2 children
Before I looked at the answers, I got 4 of the 5 right. The one I got wrong was the 4th one, where I also got confused on the writing and got the ages of 27 for John and 18 for Mary, because when Mary becomes 27, John becomes 36, in which 27 + 36 = 63. I got the ages 27 and 18 in the wrong thought, though. The last one was the quickest for me…I don’t get it. I narrowed it down to “2, 2, 9” & “3, 3, 4” since there must have been twins by my logic. I then stuck with “2, 2, 9” since before the 3 & 3 were born, the eldest would be 1, and I wouldn’t give chocolate chip cookies to a baby. I don’t know why that took me the least time, but it did.
I found the first four to be straightforward. I saw the fifth problem before. There is an alternate form where you are told that the sum of their ages is 13 and the product of their ages is my house number. There are more possibilities to go through but curiously, 1 6 6 and 2 2 9 are only two triplets summing to 13 that have the same product -- the same two triplets that appear in the solution to the version in this video. Again, the last step is to realize that the eldest child has a distinct age and hence, the solution is 2, 2, 9.
She was born on leap day, February 29th. She is 13 old, but always has a “birthday” each year regardless of whether the year has a Feb 29 or not. At 13, she has only seen 3 leap years, it will tally to 16 next year.
A year before the census taker visited, the mathematician who lived at house #13 had a five-year-old son named Adam. The family then adopted a 5-year-old named Betty (1 week younger than Adam) and a 3-month old baby named Chris. Adam was excited to have siblings! By the time the census-taker visited a year later, Adam and Betty were 6, and Chris was 1. The poor census-taker mistakenly thought the children were ages 9, 2, and 2, due to forgetting that adoption is a thing.
Puzzle 1: It's like a vantage point. From where you're standing you see many. It's the way this is phrased and the Dec 31st birthday. Selecting to talk from the 1st of January when the person has already left behind the age of 13 they get to gain a total of 3 years. And like a vantage point you see the one that has passed, one coming the same year and the one mentioned for next year. Thus from where they 're standing 3 in total. BUT come next year they WILL still be 15 years old until almost all the year has passed, almost all the months. They will STILL be 15 until the LAST day of the last month, 31st of December, and THEN and only then will they become 16 years old. Thus phrased, it still IS next year (the birthday) but not until the last day will they be celebrating the 16th. So 'they get off on a technicality' due to the position their talking from (vantage point) and the specific day of birth (Dec 31st). I elaborated a bit. Nicely explained Presh.
I solved problem 4 in a slightly different way. After I found the relationship between John's age and Mary's age, I continued as in the video, but I found the relationship like this: J is John's current age, M is Mary's current age, j is John's age when he was Mary's current age and m is Mary's age when John's age was her current age. So, according to the data, J = 2m and j = M. However, J and M are the siblings' current ages, and j and m are their ages at a certain moment in the past, so their age difference would be the same. So J - M = j - m. Now we can substitute for j and m (J = 2m m = J/2) so J - M = M - J/2. Add J + M on both sides and you get 2J = M + J/2 4J/2 - J/2 = M M = 3J/4. Then in 3J - M = 63 we substitute: 3J - 3J/4 = 63 12J/4 - 3J/4 = 63 9J/4 = 63 J = 28. Then we can find Mary's age substituting John's age in one of the equations. Nice one!
What if the eldest who is 6 was born in January, the middle child who also is 6 was born in December of the same year and their baby sibling is one? The eldest child being excited to have siblings could very easily have been a jubilant 11 month-old welcoming a newborn sibling into it’s life then 5 years later on welcoming another sibling. Stating that one child is the eldest does not infer a greater number of years, only days!
The first one. I remember this happening. A work colleague told us he texted his friend at a minute pat 12 on January 1st telling his friend that next year he would be 50 next year! Obviously his friend had his 48th birthday the day before!
Number one could have been even better by double layering the riddle. Consider all the stuff you said, but in addition consider how the phrasing in the original isn't "I will turn 16", it's "I will have my 16th birthday". Well when you turn 16, that means you have 'had' 17 total birthdays including the day you were born. So the riddle could have been pushed even one year further while still being a perfectly logical statement "Two days ago I was 13 years old, but next year I will have my 17th birthday.""
_Birthday_ means the *anniversary* of the day on which you were born. The very day on which you were born is not an anniversary of that birth, hence it's not a "birthday".
Puzzle 4 : Like you, Presh, I got this one wrong. I wish you had told us exactly how you went wrong, just in case my mistake was the same. Sometimes we learn more from our mistakes than we do from understanding the correct solution. Oh yes! One more thing....... "HAPPY NEW YEAR!" to you and all your readers.🥳🥳🥳
#4 Mary's age today represents 2X what John's age when Mary was X. Age difference is X which makes John's age today 3X. When Mary is 3X John will be 4X which sums to 7X. For 7X =63 the age difference is 9. Today Mary is 2X or 18 and John 3X or 27
14:08 "If the house number was anything besides 13, the census taker would be able to solve for the ages." Not true. If the house number was 21, the kids' ages could have been {1 , 2 , 18} , but also {½ , 4½ , 16} ; which means the census taker wouldn't have been able to solve for the ages at that point.
@mb-3faze I also got {1½ , 1½, 16} , in case the house number is 19 but this is then not a doubled house number. Other fractional triples with product 36 and that lead to an integer sum: {37.5 , 3.2 , 0.3}, with house number 41 {⅓ , ⅔ , 162} (with house number 163) but this is not realistic { 98/5 , 50/7 , 9/35 } (house number 27) { 196/5 , 75/7 , 3/35 } (house number 50) {a*p²/q , b*q²/p , c/(pq)} with positive integers p and q such that {p, q, 6} are pair-wise co-prime*, and a, b, c are positive integers such that abc = 36 and (a*p³ + b*q³ + c) is a multiple of pq . There are probably many more. [*] By saying "{p, q, 6} are pair-wise co-prime", I mean: gcd(p,q) = 1 & gcd(p,6) = 1 & gcd(q,6) = 1
@mb-3faze Other fractional triples include: {1½ , 1½, 16} with house number 19 {37.5 , 3.2 , 0.3} with house number 41 {⅓ , ⅔ , 162} (house number 163) but this is not realistic { 98/5 , 50/7 , 9/35 } (house number 27) { 196/5 , 75/7 , 3/35 } (house number 50) but these are then not doubled house numbers.
@@mb-3faze I posted two replies with other (positive) fractional triples whose product is 36 and whose sum is an integer, but for some reason youtube hides these replies (unless the comment section is sorted in "Newest first" order instead of "Most popular first" order). Three of the triples are {1½ , 1½, 16} , {37.5 , 3.2 , 0.3} and { 98/5 , 50/7 , 9/35 } .
If you allow that their are colonies on the moon, which is already fiction, then how does this affect their age anyway? And how could them being "smart" affect their age?
The moon rotates in the same time it takes to spin, which means a moon day is equivalent to a moon year (and since both are about 28 days, a 15 year old would instead only be about a 14 month old in earth-time development
@@Altemeous If a human is on the moon they are still counting time the exact same as on earth. A year is the time it takes for EARTH to travel around the sun. Other planets or moons dont have their own unique years. We measure their individual orbits in our own earth years.
the father's age problem can be solved analytically by comparing the coeficients of -9 and 10(x-2y)+y-2x. the statement -9 = 10(x-2y)+y-2x follows directly from 10x-10(2y)-2x+18 - 9 = 0, which follows directly from 10x+y-9 = 2(10y+x-9). we conclude from the statement -9 = 10(x-2y)+y-2x that x - 2y = 0 and y - 2x = 9. this implies that y - 2(2y) = - 9, which implies -3y=-9. hence y = 3. x=2y implies that x = 6. hence the father's age is 63
For the last one i solved it by the guy saying his eldest son was excited to have siblings, meaning plural, meaning twins since he knew he would get more then 1
12:25? Why did you just assume that there were twins? I know I took the same assumption, but you didn't specify this assumption before casually mentioning it in an other assumption
Because in cases of no twins the census maker would be able to know the ages of the mathematician children. The two cases of the twin siblings are the two only cases where the Third hint that the mathematician gave to the census maker is necessary.
You can't have 2 birthdays in the same year. If the current year is year 1 then you need to turn 16 in year 2, turn 15 in year 1, and turn 14 in year 0. If you were 13 years old 2 days ago and you're in year 1 that means that the only day that your birthday could be is December 31. If it's before 31.12 then you'd currently be in year 0. If it was after, then your 14th birthday would be in year 1.
I'm not going to exactly remember the setup or the specific numbers in the original but the best age riddle I ever heard was something crazy like: Jimmy's age in 5 years will be the same as if you took his mother's age, subtracted 3, halved it, subtracted 3 again, and halved it again. If Jimmy's mother is 26, where is the father? In a way, this one is solvable 😂 I gave part of it away by implying that the numbers matter. No misdirection there 🧠
Spoiler - answer ahead: Most sane people will look at it and dismiss it as nonsense, but then you start to work it out and you find that Jimmy's age works out to be -¾. Huh. Well then something must have gone wrong with the writing of the riddle. According to the math, Jimmy won't even be born for another 9 mon- 😮
@@DrEcho -- Exactly. Jimmy was either just conceived or hasn't been conceived quite yet. In the latter case, there is as of yet no father. (I'm not going to add the complication of calculating weeks instead of months for a full-term baby Jimmy.)
@@DrEcho definitely ambiguous, but if you make the blatant assumption on the average time of pregnancy being exactly 9 months then I guess you can make a joke related to how they are having sexual intercourse right at the very moment Jimmy goes into existence, im gonna guess that's the idea
@@DrEcho I thought the answer would be something like the mother being 13 yo when conceiving Jimmy, and therefore the father would be on jail. But the actual solution is more hilarious 🤣
@randomgamer-te8op most definitely there are a few big assumptions, not least the exact gestational age of Jimmy at birth. I'm guessing that the idea of the joke is once you infer and take for given that there even is a somewhat cogent solution, you can answer with a bit of certainty after you overcome the math barrier. Akin to puzzles like last problem in this video; only when you're given the instance that you have enough information to solve it, can you actually solve it. But, yes, you are correct that the alternative 'correct' answer is "the father could be literally anywhere, as we do not have the information required to make such a deduction".
Birthday Dec 31. Conversation on Jan 1.
Two days ago Dec 30 he was 13. One day ago turned 14. Today Jan 1 is a new year, will turn 15 at the end Dec 31 and be 16 at the end of the next year on the last day of the year.
I think it's cuz his birth (when he is 0 years old) is his 1st birthday and thus him turning 15 is his 16th birthday and fitting that into a coherent timeframe is trivial :)
Vincent is right,
@@A.Martinunless if it’s a leap year
Yeah, this one was pretty straightforward.
@@Criz454 That is not it. Everyone turns 1 when they are 1 and there first birthday is when they are 1 years old.
What's so "seemingly impossible" about the second one? It's very straightforward. You just count the days backwards, then forward.
It's literally just the wording is confusing
Father-son was even simpler.
THE WORDING ARE SO CONFUSING FOR ME SOMEHOW? Two day after the day before birthday. My brain gone
I'd also say the same for the first one, but what they mean by "seemingly impossible" is that you might at first think it's impossible to be 13 two days ago and to turn 16 next year... The way that it sounds is odd, until you think about it. Perhaps to some people, the second one does sound similarly "seemingly impossible"
That is the only one I got wrong, since I misread the wordings. Pretty ironic
Puzzle 3 applies to me - I am almost exactly 27 years older than my son. When he was 03 years old, I was 30. Eleven years later, when he was 14, I was 41, and the pattern re-occurs every eleven years. I have thought a lot about this over the years, and it turns out that it will only work when the difference in ages is a multiple of 9, and it will always then repeat every eleven years.
Two days ago i was 13 (december 30) the very next day i turned 14 (december 31). Today is january 1 (it literally is) this year i'll turn 15 then next year ill turn 16. December 31 is my birthday. Happy new year!
13 years and 364 days isn’t 13 years. If you allow approximation, it’s 14 years
@@blackquintet Except no one says “I’m 13.99 years old” the day before their 14th birthday. Age is measured as an integer.
@@kurisueru 13.5 years is thirteen and a half year. Some people might say 13 years 6 months. 13.99 is 14 years. But maybe you live on a different planet.
@@blackquintet Uh huh. And if you walk into a liquor store in the US saying you’re 20 years, 11 months, and 29 days old, the clerk will tell you to come back in a couple of days when you turn 21. Though I should have been more specific and said that age is expressed as an integer instead of measured. People that regularly walk around giving their age the way you describe are the ones often treated as though they’re from another planet.
On puzzle 3, I figured it out almost instantly because a few years prior I figured out that when you reverse a 2 digit number the difference is a multiple of 9
Whenever someone starts with
"When I was your age..." again,
I'm gonna throw them off with
"Yes, I was your age less than twice my age back then."
and leave them wondering.
Legitimately wondering, cos it doesn't make sense.
@@peteneville698 Watch how he solved the 4th question again...
If M is my age and S is someone's age, when that someone was M years old, i.e. S - M years ago...
I would have been M - (S - M) = 2M - S years old back then.
The only thing I can see relevant about the cookies is that the conversation takes place after 1938.😅
Here's a simple solution I did for puzzle 4:
Lets say that d is the age difference. Mary's age d years ago is half of John's current age and since the age difference itself is d, 2d is half of John's age, so d is 0.25 of John's age
The sum of current ages is 1.75 Johns, and d years later, both ages will be 0.25 Johns higher, so it will be 2.25 Johns. Divide 63 by 2.25 to get 28, 28 * 0.75 = 21.
Using johns instead of letter. Idk this seem american ways lmao. Sorry for that
that's what I did.
For problem 4, it’s easier if you define an additional variable, d=j-m, and solve in terms of d.
yup, I used delta like you.
2:05 This 2nd puzzle is based on Married with Children, when Al Bundy was in a rare happy moment, and told his wife,
"Next to the day Before I met you, This is the happiest day of my life!"
True genius right here (no sarcasm)
@@TruthTF Haha! I just saw a similar episode yesterday when Kelly got a job and Al told her,
"This is the proudest moment of my life.
And it shall remain so, until the day your mother leaves me."
Would any 14-year-old have the credentials (academic or otherwise) to be called a “logician”? Seems a stretch to use that title on someone that age. It would have made more sense to add, say, 30 years to each of the numbers in the problem.
Doogie Howser begs to differ.
logician is as logician does.
"logician" is not a degree or academic title, as far as I know.
@@tomriddle8933 Ah, Doogie Howser! He is Legen... Wait For It... Dary!
;-)
A logician is a technician so no degree is required. I was taught logic at highschool.
NOTE: unless I say otherwise, I did not use the video to answer these questions.
Puzzle 1
Birthday is on December 31st, and the riddle is on January 1st. 2 days ago, they were 13 (Dec 30)
The next day, they turn 14 (Dec 31)
The next day is new year, so this year they turn 15 and next year they turn 16.
Puzzle 2
Yesterday = Wednesday
Day before = Tuesday
Day before = Monday
2 days before = Saturday
1 day after = Sunday
Birthday was on Sunday.
Puzzle 3
x = son’s age
Age difference = x - 9
Assuming there is no illegal situation here, the dad should be at least 18 when the son was born. We can use trial and improvement to find the exact age.
18 + 9 = 27 (dad would be 72, and 27 and 72 are not 18 apart)
Estimation: 35 - 9 = 26 (dad would be 53, and 35 and 53 are not 26 apart)
Estimation: 36 - 9 = 27 (dad would be 63, and 36 and 63 ARE 27 apart.)
Using maths and trial and improvement, we have figured out that the son is currently 36 and the dad is 63.
Puzzle 4
John’s age = x
Age difference between John & Mary = y
Let’s rewrite the information so it makes more sense (or less sense if you don’t understand algebra lol)
x = 2x - 4y
2x + y = 63
From the top equation, we can work out that x = 4y. This means that the bottom equation can be rewritten as:
8y + y = 63
Gather like terms:
9y = 63
Divide both sides by 9:
y = 7
x = 4y = 4*7 = 28
28 - 7 = 21
John is 28, Mary is 21.
Puzzle 5
I admit that for this one I watched part of the video, but only saw up to 12:07.
You can infer that the census taker knows the house number, therefore there must be multiple ways to add up the children’s ages to get the house number and still get 36 when multiplying them.
36 as a product of prime factors is: 2*2*3*3.
There are 6 possible age combinations for 3 children from this:
1, 1, 36 (sum 38, also unlikely as they are referred to as “children”, and a 36 yr old is an adult)
1, 4, 9 (sum 14)
1, 6, 6 (sum 13)
2, 2, 9 (sum 13)
2, 3, 6 (sum 11)
3, 3, 4 (sum 10)
Out of these, only 2 share the same sum. Both combinations 1,6,6 and 2,2,9 have a sum of 13.
However, as the mathematician refers to his eldest *child* (not children) and how he was excited to have *siblings* (not a sibling), we can infer that he is a good few years older than his siblings. As 1,6,6 would involve him having a twin, this combination is impossible, therefore, the eldest is 9 and the other 2 children are both 2 years old.
Riddle 5: Census Taker: Um, I have a long line of houses in which to take census after yours, and if they all give me riddles like this, I'll never get my job done.
2:36 nice reference
Is the next birthday when he turns 64?
Now it looks as though they're here to stay...
Now I need a place to hide away...
an incorrect one :D
What reference
It's quite nice seeing ukmt challenges, my secondary school offered them to me for 2 years in a row, and a boy who was in my class was the best in the whole south west!
I do the UKMT maths challenge every year at my school (only top set do it though) and last year I got through to the 2nd round (the only one in my year to do so). I didn’t get any further though
@Hi-rw8vr well done! They can be really hard. Which one did you do, junior or intermediate?
I did Puzzle 4 a bit differently. After working out John was 4/3 Mary's age, I realized that the sum was 9 "differences in age" - John is currently 4 "differences" old, Mary is 3 "differences" old, then you need to add one more for each of them the time Mary spends catching up to John's current age.
63 divided by 9 is 7, so that's the difference in their ages. This means John is currently four 7s old, making him 28, and Mary is currently three 7s old, making her 21.
That's insane, i need some sleep
I just divided 63 by 3, because I "know" Mary's current age has to be a third of the final sum. Then I double-checked that 14 was half of 28 and that 28+35=63.
That "know" was a source of endless frustration for my teachers throughout the years. Certain things in word problems just seemed obvious to me, and whenever I tried to use formulae to show what I knew, I inevitably got the wrong answer.
The 5th puzzle is truly an all time classic, presented in different versions. There's a channel called Ted-Ed where they present riddles and the same riddle was presented in a life and death situation, where if you can't find a passcode, you'll be thrown into a cage of hungry salamanders😅
The last one had me stumped. Great video!
I'm glad you added the welcoming having siblings bit because just liking cookies really doesn't mean they have a unique age since there's always an elder among twins. But welcoming having siblings still isn't great, since it also means they'd need to be old enough to speak etc. when their siblings were born, as well as the fact that it can be read to imply they were welcoming multiple siblings at once (if I welcomed two siblings individually then at no point did I ever welcome siblings plural).
The eldest would not have been able to "welcome" siblings if they were a twin, since functionally they'd always be around one another. You can't be excited about gaining something you already possess.
you can welcome having siblings through gestures, not necessarily speaking, and you cant say in the pazzle "he is not old enough to talk" anyways so it doesnt matter.
when someone says he welcomed having siblings doesn't mean they are twins, it's not a quote "thank you for my siblings", it could have been individually.
@@mydudestudios6244 you understood what he said wrong, read it again
@@shakedeilstein2448 o for sure, I just meant it's still open to misinterpretation and feels a bit clunky. It's definitely an improvement on just saying the eldest likes something or other though!
The phrase in the riddle actually was “he was excited to have siblings” as in they were aware of not having siblings and looked forward to having some. Twins or other multiples would never be aware of not having siblings, as mydudestudios6244 stated. “Welcoming having siblings” would still refer to looking forward to having future siblings, not literally greeting them in any manner.
December 31st is the asker's birthday. The current day is January 1st, so two days ago was December 30th, the last day the asker was 13. The asker will be 14 for the entirety for the entirety for the current year, EXCEPT for the last day, December 31st, when they will be 15. They will then be 15 for all but the last day of the NEXT year, when they will turn 16.
You need to edit and delete an extra phrase "for the entirety." In case you don't know, click the three dots to the right of your post.
You still never explain what “the eldest child loves chocolate chip cookies” has to do with solving the problem! If the mathematician had only said, “the eldest child was excited to have siblings”, without ever mentioning his son’s proclivity for cookies, the census taker would still have been able to derive the answer. From that statement alone, it is clear that the eldest child remembers a time in his life when he had no siblings-which would not have been the case if the kids were 6-6-1. Therefore, the only possibility is 9-2-2-regardless of whether any of the kids like cookies.
I think the important part there was the "My eldest child" refering to only 1 being the oldest and "he was excited to have siblings" referring to the youngest being twins (maybe)
Yeah the info about cookies is irrelevant. It has nothing to do with solving the problem.
No, saying you were excited to have siblings and remembering a time when you didn't doesn't tell you anything because that would have to always be the case. Of course there was a time when the oldest child was the only child, how could that ever not be the case?
verkuilb It's because the "excited to have siblings" part was added in later. To begin with, you were supposed to get the information that the oldest child isn't a twin from the phrase "eldest child" since most people don't think of one twin as being older than the other. Saying "loves chocolate chip cookies" was just an excuse to say SOMETHING about the oldest child. Then it was pointed out that not only could the oldest be a twin, but also he could be less than a year older than the next sibling, so they could be the same age without being twins. For example, my ex's two brothers were born in the same year, one in January and the other in December. So the detail about "excited to have siblings" was added in, indicating that the oldest would have to be over 1 year old when the next child was born. You're correct, this makes the chocolate chip cookies irrelevant, but whoever made up this version of the question didn't realize it and left it in.
It means that eldest is above age of 4 as the teeth of 4 year olds get spoiled too quickly
I don't like that most of those are just confusing wording and not really hard puzzles otherwise
Greek jumpscare
@Sior-person My username? It's simply a name and the word "Website" in Greek. No math involved (this time)
@ΝίκοςΙστοσελίδα yes yes I know I know
It’s just-
Not every day you see Greek in a TH-cam comment section 😭😭
Μπράβο ρε διάνοια πατριώτη!Εσύ πήρες όλο το 2ο Πεδίο κι'έφυγες για MIT!Εμείς τα "ταπεινά χαμομηλάκια" του 1ου Πεδίου,θα πάμε να κλάψουμε στην ορθή γωνία μας...
@@despobordello whar
Τι-
(Paused) I had a rough time parsing puzzle number 4 but I think I have the answer.
John today is twice Mary's age when John was Mary's age. So if we say J is the age of John and M is the age of Mary, then what this line says is that when John was the age Mary is today, Mary's age would have been half the age he is now. So J = 2(M - (J-M)) which if you solve for J yields 4M/3 like so:
J = 2(M - (J-M))
--> J = 2M - 2(J-M)
--> J = 2M -2J + 2M
--> J = 4M - 2J
--> 3J = 4M
--> J = 4M/3
When Mary will be John's age, the sum of their ages will be 63. To get Mary to John's age you would be adding the difference in their ages to her age, but John will also age the same amount in the same time frame, so you must add that difference to his age too. So 63 = (M + (J-M)) + (J + (J-M)). But we now know J is 4M/3 so I can just substitute that in for J and solve for M.
63 = (M + (J-M)) + (J + (J-M))
--> 63 = (M + ((4M/3)-M)) + ((4M/3) + ((4M/3)-M))
--> 63 = -M + 3(4M/3)
--> 63 = -M + 4M
--> 63 = 3M
--> 21 = M
So Mary is 21. And since we know J = 4M/3, then J=4(21)/3, or 28. So John is 28 and Mary is 21. When John was 21, Mary was 14 so he is twice that age now. When Mary is 28 John will be 35 and 28+35 does indeed = 63.
Now to unpause and see if I got the right answer...
5:20 - We also know that the x > y (because the father's age 10x + y is greater than the son's age 10y + x), and we know that y is odd, because the father's age - 9 years is even (2 times the son's age - 9). We can also re-arrange 8x + 9 = 19y to 9 = 19y - 8x = 18y - 9x + y + x, which tells us that 9 = 9 (2y - x) + (x+y), which tells us that x+y must equal 9.
All three constraints mean the father's age must be 81 or 63. From there it's trivial to see that 63 is the father's age.
Where in the riddle did it say the ages had to be two digit numbers?
@@jacobgoldman5780
1. The father's age is a 1-digit number.
If the father is younger than 10, then the father is the same age as the son. This is an impossibility.
2. The father's age is a 3-digit number.
The hundreds digit of the father's age must be greater than the ones digit, so the ones digit must be a 0; if the ones digit is a 1, the hundreds digit must at least be a 2, and the father would be 200+ years old. But the ones digit of the father's age must be odd, because the father's age 9 years ago was double the son's age 9 years ago. This is a contradiction: the ones digit of the father's age cannot be 0 *and* an odd number.
@@jacobgoldman5780 The only ages that work are:
1. The father and son are both 9 years old;
2. The father's age is 6 * 10^n + 3, the son's age is 3 * 10^n + 6. Examples: 63 & 36, 603 & 306, 6003 & 3006.
The only pair that makes sense is 63 & 36.
I solved the last puzzle a bit theoretically:
I wasn't really familiar with the concept of house numbers, so I skipped it (which kind of made my reasoning flawed).
The eldest child had a distinct age, so it couldn't be 6,6,1.
The child also loved choco-chip cookies, so it probably had well-grown teeth and hence it couldn't be 6,3,2; 4,3,3.
The eldest was also excited to have siblings; so the other two could be twins.
That left me with: 36,1,1; 9,2,2.
Now 36,1,1 seems a bit far-fetched, so their ages are probably 9,2,2.
The twin assumption is probably the greatest flaw, but the wording made me feel so.
10:49 What was your mistake?
If you misinterpret the odd framing of the question,it can imply they are the same age, so they would be 31.5 years old. Many people do this and I remember in my math club in elementary school this same problem became a point of argument.
You can read it as when John was Mary's age now, for example if John is 40 and Mary is 20 now then John is twice Mary's age now. Wording is not great but you can tell this is not the answer as the math doesn't work.
I keep interpreting it as, You are 13 and I am twice your age which makes me 26, when you are 26 I will be 39 and add them together it's 65
I know this isn't 63, but if you start with 12 the answer is 60
I'm still confused
@@matthewboyd8689if you reword it as John's age today is twice Mary's age at the time john was Mary's age, does that make it any easier to read?
Puzzle 1, you should pick the most extreme case possible to speculate. December 31 birthday, January 1st day of announcement. December 30th was 13. December 31, becomes 14. This year, 15. Next year 16.
Interesting that you posted the logician birthday riddle right before the New Year, perhaps priming us for that answer.
The census taker has some talents to work that out in his/her head!
3:31 i think i might've had this one as a bonus question in one of my math classes
so for puzzle 4: you are working with differences of ages, it makes total sense to declare some variable to use for it. Like d = J - M (as John is the older one)
We know that M = J - d and J = 2(M - d), so J = 4d and M = 3d
We also know that J (Mary's age when she is as old as John is today) + (J + d) (John's age at that time) = 63
so 2J + d = 63 -> 2(4d) + d = 63 -> 9d = 63 -> d = 7
which means J = 4*7 = 28
and M = 3d = 3*7 = 21
Great video dude! Keep up the great work 👍
Puzzle 4: I think it's less confusing to keep the first equation as 3J=4M and to substitute for 3J instead of J in the equation 3J-M=63. That way you directly have 4M-M=63. It's pretty much the same thing but deviding in equations like this always confuses me.
Great video btw (:
For the fourth riddle, I found it easier to put the past ages at J and M and the present ages as J+x and M+x, making the future ages J+2x and M+2x. In the end, I had converted everything to xs and just had to divide 63 by 9 (the number of xs) to get to x=7.
15:04
2:35: 137 points to you for the Beatles reference.
0:30 i figured it out in like 30 seconds its december 31st
For the first riddle, although your answer is probably the correct solution. I read the riddle differently. I would argue you become 15 years old when you have your 16th birthday. Your first birthday is technically the day of your birth. 1 year later you turn 1 year old by having your 2nd birthday, it's just the 1st anniversary of your birth. Though you can get more technical about it and argue you become 15 years and 9-10 months old on your 16th birthday.
Your answer is correct. I just think it's funny to think of it like this.
For puzzle 3, you should have also shown that nine years earlier the father's age (54) was twice the son's age (27).
I think that 4th puzzle could definitely be worded better, specifically the first sentence, it should be "John today is twice the age that Mary was when John was Mary's age"
I spent a long time getting John and Mary wrong! Happy I got the others though, the census one is good but needs a bit of assumption that the visitor is a genius.
Thumbnail/First puzzle: December 31. If it is January 1, the girl was 13 on December 30. She turns 14 on December 31 and is still 14 on January 1. Next year, she'll be 16 because her birthday will have already passed again (making her 15) by the time the new year rolls aground. Before that year ends, she'll be 16.
Second puzzle: Sunday. The day before the day before yesterday is three days before today. Today is Thursday, so three days prior is Monday. Monday was two days after the day before the birthday. Monday is two days after Saturday, so Saturday was the day before the birthday, thus the birthday is Sunday. This wasn't wasn't really a puzzle as much as it was just parsing through and counting the number of "day before"s.
Third puzzle: The father is 63 and the son is 36. The father and son can only have so many ages, such as 91 and 19, 92 and 29, etc., but most will not fit the rule for the doubled age. Any triple-digit ages can be disregarded for this reason. Realistically, the easiest way of solving this is to brute-force it with Excel or Google Sheets, using formulas to fill it out. For example, A1 is 1, B1 is A1*2, C1 is A1+9, D1 is B1+9, and E1 is FLOOR(D1 / 10, 1) + (MOD(D1, 10) * 10) - C1. From there, autofill the spreadsheet. If E equals 0, there's the answer. Starting at 1, the answer will be on the 27th row.
Fourth puzzle: John is 28 and Mary is 21. I'll get to the lengthy math after the fifth puzzle.
Fifth puzzle: Their ages are 2, 2, and 9. I'm pretty sure you've done a video on this one before, but there were arguments in the comments about how "my oldest loves chocolate chip cookies" does not disqualify being a twin. There are only so many combinations of three numbers with a product of 36. Of them, their sums will produce a list of possible house numbers. The census taker knows the house number. If the correct house number appeared only once in the list, the census taker would know the answer. Since they did not, the house number must have appeared more than once. I broke out the spreadsheet again to help simplify finding this. The only sum that appears more than once is 13, using the ages (1, 6, 6) or (2, 2, 9). The cookie hint indicates there is an oldest child that does not have a twin, therefore it must be the latter option.
With that, I got all five puzzles.
Fourth puzzle explained: I do not like the wording on this one, but I think this is what it is saying:
John is older than Mary.
John's current age is an even number.
When Mary age was half John's current age, John's age then is Mary's age now.
When Mary reaches John's current age, their ages will have a sum of 63.
Jf = John's future age for a sum of 63
Mf = Mary's future age for a sum of 63
J = John's current age
M = Mary's current age
Jp = John's past age
Mp = Mary's past age
63 = Jf + Mf
Mf = J
M = Jp
J = 2 * Mp
Jf - J = Mf - M
J - Jp = M - Mp
J - Jp = M - Mp
J - M = M - Mp
J = 2 * M - Mp
2 * Mp = 2 * M - Mp
3 * Mp = 2 * M
1.5 * Mp = M
63 = Jf + Mf
63 - J = Jf - J + Mf
63 - J = Mf - M + Mf
63 - J = 2(Mf) - M
63 - J = 2(J) - M
63 - (2 * Mp) = 2(2 * Mp) - M
63 = 3(2 * Mp) - M
63 = 3(2 * Mp) - 1.5 * Mp
63 = Mp * (6 - 1.5)
63 = 4.5 * Mp
14 = Mp
J = 2 * Mp
J = 2 * 14
J = 28
M = Jp
J - Jp = M - Mp
28 - M = M - 14
42 = 2 * M
M = 21
I have never once heard the word distinct to describe an age. It's pretty much never used for that. Distinct means something stands out, is definite, recognizable, as if it's been highlighted somehow. I think the word you're looking for is unique.
Last is fabulous 👌
Nice fix to puzzle 5.
I solved nos 4 and 5 differently.
For 4, I realized that when Mary is John's age today the sum of their ages would be 63 and that means 3J-M=63. So, I just made a 3 column table and found with the columns "J", "M" (calculated as 3J-63), and "M2" (which was Mary's age when John was Mary's current age. calculated as 2M-J). And I just had to go down until I saw an answer where J is double M2.
J - M - M2
31 - 30 - 29
30 - 27 - 24
29 - 24 - 19
28 - 21 - 14 And there was my answer.
For 5 I'm not sure this was a valid solution, but I got the same answer as Presh. I realized that the answer had to be something where the oldest child had a different age than the other siblings, because they were excited to have siblings, which means there's a time they can remember where they didn't. I also figured the two younger had to be twins, because they were excited to have siblings (plural). So, I figured the only possibilities were 1, 1, 36; 2, 2, 9; and 3, 3, 4. I figured a 1 year old would be too young to comprehend the fact that their father's baby mama was pregnant and get excited. And I figured that "loves chocolate chip cookies" isn't the first thing you'd mention about a 36 year old. So, I got 2, 2, 9.
So, if you wanted to do it completely algebraically without using the table method you would realized J=2(2M-J), which means 3J=4M and 3/4J=M. Also, J+M=63 So, 7/4J=63. Therefore J=28.
8:21 I was wondering why you brought that three over because you just made Mary's age more complicated.
That’s what I thought. 3J=4M actually ends up working perfectly for the second half.
For puzzle 3 I had to write a bit of code to check stuff before I found it, then I saw you doing the equations and thought "Oh there's gunna be some easy way to do this" and then you pull out the spreadsheet and check every option just like I did.
For the final puzzle, we're assuming the census taker is a mathematician as well. His response "I don't know" tells me nothing, as that's probably the answer of any census taker when given such an absurd answer.
The fact that the census taker says "Ah, now I know" after the mathematician started babbling about how the eldest loves chocolate chip cookies, indicates that the census taker is also a math aficionado.
@yurenchu Haha or he jotted down on his notepad "Crazy person loves here" and got out of there as soon as possible.
Of course you're right, just being facetious.
@@stephenj9470 Haha, exactly, "don't touch the tea". :-D
Puzzle three be like: *you didn't have to cut me off, make out like it never happened and that we were nothing...*
the second last one was crazy ngl
For puzzle 3 I did it this way:
F is the father’s age and S is the son’s age
F - 9 = 2(S-9)
Which simplifies to S = (F+9)/2
Then you can graph it and find a value where S and F are the same number only reversed.
There are other ways to do it too, but this was what came to my mind first for some reason.
today is the logician's birthday! happy birthday
For puzzle 4, I had more variables, one for each age in each timeframe (past, present, future), and more systems of equation. I managed to rearange and substitute through to isolate John's age, then solve for Mary's age.
The first puzzle I interpreted differently. Based on the wording, the logician is 14 years old/14 years of age presently being she either is still 13 and mentioning she was also 13 years old two days ago, or she is 14 years old presently having a birthday over the past two days. She has had 15 birthdays once reaching 14 years of age (including the day she was born, 0 years old), so next year when she turns 15 years old it will also be her 16th birthday.
If the eldest child were 6 years 11 months old and the second-eldest 6 years 0 months old, the eldest child wouldn’t have learned to articulate such a complex thought as “I’m excited to have siblings” at less than a year old. That’s the new assumption
For the fifth puzzle, I don't believe knowing that there is an oldest child means knowing that the oldest and second-oldest children have different integer ages. I am an uncle to two sisters who were born ten months apart, so for two months every year, they have the same integer age as one another, but there is clearly one who is older and one who is younger. Even in the more common example of twins, one baby generally arrives before the other one, even if it's only by a matter of minutes.
The only scenario I could imagine is if the twins are born through C-section, and somehow both babies were extracted at the same time. I don't know enough about that process to know if that is even possible, but if it's possible, then there would be no oldest child, but two oldest children.
I knew twins born via C section growing up. They knew which of them was taken out first.
I think the relevant bit is actually that "he was excited to have siblings." Siblings, plural, as in twins. If they were only expecting 1 child, then he'd only have been excited to have **a** sibling.
Saying he was excited to have siblings does not mean that they arrived at the same time. When he was an only child, he could have said, “I can’t wait to have brothers and sisters”, without knowing if/when they might arrive, or in what quantities. However, the fact that he was excited to welcome siblings DOES indicate that there was a time when h he was an only child, and was old enough to be aware of that fact.
I know a crapload of twins, and you know what? no one cares, although they know, and it is not relevant.
Incidentally birthday and birth anniversary can be a confusing concept.When you are 1 year old,it is your second birthday.
It isnt. This has nothing to do with the puzzle. When you turn 1, it is your FIRST birthday. Birthday means birthday anniversary - not the literal day AND year you were born. Noone has ever celebrated a newborn babies birthday of the 0 years of life milestone.
@@NowhereMan7 Exactly. On the 21st anniversary of one’s birth, no one goes out with friends to celebrate their “22nd birthday.”
Strictly, you are 21 on the day before what is commonly called your 21st birthday: so; if you were born on 21st June 2004 the 21st June 2025 is the 22nd time you have been alive on 21st June- you had been alive 20 years and 365 days immediately after whatever time you were born on 20th June.
Birthdays are the years weve finished. If you say "Im 21." that means youve finished 21 years, and started your 22nd. On the day of your birth is your 0th birthday bc youve finished 0 years and are starting the 1st.
@@rogerdines6244 What you are doing is to force a poppy to a thread. You are wrong!
The second puzzle is a lot easier for russian language speakers, because we have separate names for the "the day before yesterday" and "the day after tomorrow"
The first puzzle has a second solution (actually I would argue this is the _correct_ solution): your 1st birthday is the day you are born, when you are 0 years old. Thus, your Xth birthday is the day you turn X - 1 years old. Therefore your 16th birthday is the day you turn 15. It's the same as how the 20th century is the 1900s rather than the 2000s despite the latter being the one that actually starts with 20.
So we'd have: 2 days ago, they were 13. Then they turned 14, but it was the 15th birthday they had. Next year they'll have their 16th birthday, when they will turn 15.
Also for 5; "...he was excited to have siblings" suggests that twins were arriving (rather than having "a sibling" or "another sibling") so the younger two would be the same age?
Well, since (half) my post got deleted, I'm just gonna paraphrase, because I'm lazy. Back in the '90s, when I was a sophomore in High School, my math teacher somehow divined that I was bored to death with my Algebra II classwork, and approached me privately with an almost exact copy of the Census Taker problem you laid out here (the man was in his 90s, to allow for the possibility of a 72-year-old child, and the "oldest child" was limited to liking chocolate cake; the "enjoyed having siblings" remark was simply unnecessary, as it probably is here). My immediate reaction to reading the problem was, 'I can't possibly solve this, there isn't enough information." But, my teacher said that there actually was, so I basically brute-forced the issue and had the correct answer within twenty minutes. Still, to this day, I love the way the problem is constructed, since the census-taker's ignorance is obviously important but comes across as being so pointless.
A few years later, this very problem inspired me to construct my own "unsolvable" problem that is totally solvable and yet, unlike the Census Taker problem that only _seems_ impossible at first glace, actually seems completely unsolvable from any possible perspective, and I did it by including, as one of if not _the_ core aspect of the problem, a premise that seems on its face to be plainly and incontrovertibly just flat-out _wrong._
I came up with this problem three decades ago. I don't know how many people have read it, but I posted it back on Slashdot, TotalFark, SomethingAwful, the math and physics forums on Yahoo Answers, etc. Not only has no one, as of this very day of January 5th, 2025, ever gotten the correct answer, I don't think I've ever seen anyone employ the sort of approach that could even _lead_ to the correct answer. Now, there _is_ a correct answer, and despite giving zero hints when I first posted it, I _will_ save a _lot_ of time by letting you know that neither zero nor infinity are the answer, or even _appear_ in the answer. The answer is a positive, finite value, and that's all the clues you're getting. Now, since my original post screwed up, I'm gonna try posting this part first, and _then_ edit the actual question in more or less from unchanged from my original document. Oh, one more hint: this question employs geometry, but it is a purely _mathematical_ question. Physics are _involved_ in the question, but they're not _relevant_ to the question, nor are they to the answer. This is pure math. Here we go:
-=-=-=-=-=-=-=-=-=-=-=-=-=-
*The Flat Sphere*
First, we must consider a hypothetical sphere of unspecified and irrelevant physical attributes. Only a few basic properties matter:
1) It is a physical object.
2) It exists somewhere in the universe.
3) It is solid, or atleast has a solid surface.
4) It is perfectly spherical.
5) Its size is dynamic (explanation below).
So, a completely round solid spherical object... a cue ball from a pool table suffices perfectly for this hypothetical object. Let us also assume that physical forces such as gravity, electromagnetism, inertia, and angular momentum are not at work, or atleast that this object is immune to them and they are irrelevant for our purposes; our cue ball is not going to be collapsing under its own gravity or breaking apart due to shearing forces. Our sphere, for that matter, is not even comprised of atoms or other particles, but is constructed merely from a solid material. While it exists somewhere in our universe, let us consider it to be indestructible.
And, finally, to the crucial element. We can take this sphere and increase or decrease its diameter, and thus its radius, circumference, surface area, volume, etc., at will. Although it really makes no difference, let's assume it begins at the size of a normal cue ball, a few inches in diameter. From here, we can manipulate it and alter its size down to the diameter of a neutrino or smaller, or up to the diameter of the Milky Way or larger. We can make it as large or as small as we possibly want. Now, here is the riddle: As the diameter of a sphere increases, its curvature decreases, and vice versa. So, what size does our object have to be in order for the surface of the sphere to become *completely flat*?
There is, in fact, despite first appearances and protests to the contrary that I have received in the past, a correct answer, and only *one* correct answer, to this riddle. In addition, *all* of the information necessary to arrive at the correct answer has been provided here, so I’m afraid no further help or details can be given. Best of luck!
For the first one it can be any day since jumping around time zones can effectively double your birthday count
Puzzle 4 sounds like the annoying employee who doesn’t want to give away his age.
I remember a puzzle similar to puzzle 5; I don't know the numbers involved or the setup - but the end of the puzzle is something like Mathematician A "I don't know" followed by Mathematician B saying "Now I know" followed by Mathematician A saying "Now I know, too!"
he did a few of this "i dont know" "oh, now i know" in this channel, but he did an even similar problem in this channel, it was pretty much the same just with 2 children
Before I looked at the answers, I got 4 of the 5 right. The one I got wrong was the 4th one, where I also got confused on the writing and got the ages of 27 for John and 18 for Mary, because when Mary becomes 27, John becomes 36, in which 27 + 36 = 63. I got the ages 27 and 18 in the wrong thought, though.
The last one was the quickest for me…I don’t get it. I narrowed it down to “2, 2, 9” & “3, 3, 4” since there must have been twins by my logic. I then stuck with “2, 2, 9” since before the 3 & 3 were born, the eldest would be 1, and I wouldn’t give chocolate chip cookies to a baby. I don’t know why that took me the least time, but it did.
I found the first four to be straightforward. I saw the fifth problem before. There is an alternate form where you are told that the sum of their ages is 13 and the product of their ages is my house number. There are more possibilities to go through but curiously, 1 6 6 and 2 2 9 are only two triplets summing to 13 that have the same product -- the same two triplets that appear in the solution to the version in this video. Again, the last step is to realize that the eldest child has a distinct age and hence, the solution is 2, 2, 9.
She was born on leap day, February 29th. She is 13 old, but always has a “birthday” each year regardless of whether the year has a Feb 29 or not. At 13, she has only seen 3 leap years, it will tally to 16 next year.
The problem with that solution is 13 isn't divisible by 4 which is how often leap years occur. So she'd have to be 12 for 4 years before turning 16.
@ 12 / 4 =3
13 + 3 =16
They merely celebrate her birthday at a later date on most leap years.
@@eds1942 But that brings you to the same problem as in the video. At best she can turn 15 the next year.
@ 16 birthdays, not years.
@@eds1942 As in she'd celebrate two birthdays each leap year?
A year before the census taker visited, the mathematician who lived at house #13 had a five-year-old son named Adam. The family then adopted a 5-year-old named Betty (1 week younger than Adam) and a 3-month old baby named Chris. Adam was excited to have siblings! By the time the census-taker visited a year later, Adam and Betty were 6, and Chris was 1. The poor census-taker mistakenly thought the children were ages 9, 2, and 2, due to forgetting that adoption is a thing.
I liked that last one
Puzzle 1: It's like a vantage point. From where you're standing you see many. It's the way this is phrased and the Dec 31st birthday. Selecting to talk from the 1st of January when the person has already left behind the age of 13 they get to gain a total of 3 years. And like a vantage point you see the one that has passed, one coming the same year and the one mentioned for next year. Thus from where they 're standing 3 in total. BUT come next year they WILL still be 15 years old until almost all the year has passed, almost all the months. They will STILL be 15 until the LAST day of the last month, 31st of December, and THEN and only then will they become 16 years old. Thus phrased, it still IS next year (the birthday) but not until the last day will they be celebrating the 16th. So 'they get off on a technicality' due to the position their talking from (vantage point) and the specific day of birth (Dec 31st). I elaborated a bit. Nicely explained Presh.
I solved problem 4 in a slightly different way. After I found the relationship between John's age and Mary's age, I continued as in the video, but I found the relationship like this: J is John's current age, M is Mary's current age, j is John's age when he was Mary's current age and m is Mary's age when John's age was her current age. So, according to the data, J = 2m and j = M. However, J and M are the siblings' current ages, and j and m are their ages at a certain moment in the past, so their age difference would be the same. So J - M = j - m. Now we can substitute for j and m (J = 2m m = J/2) so J - M = M - J/2. Add J + M on both sides and you get 2J = M + J/2 4J/2 - J/2 = M M = 3J/4. Then in 3J - M = 63 we substitute: 3J - 3J/4 = 63 12J/4 - 3J/4 = 63 9J/4 = 63 J = 28. Then we can find Mary's age substituting John's age in one of the equations. Nice one!
Happy birthday, unnamed logician friend!
For puzzle # 2 my own birthday was actually on a Sunday ! What an unbelievable coincidence.
What if the eldest who is 6 was born in January, the middle child who also is 6 was born in December of the same year and their baby sibling is one? The eldest child being excited to have siblings could very easily have been a jubilant 11 month-old welcoming a newborn sibling into it’s life then 5 years later on welcoming another sibling. Stating that one child is the eldest does not infer a greater number of years, only days!
2:35 - I see what you did there… 😶🌫️
i got all of them. First time for ages
The first two puzzles seemed super easy to me.
Got all 5 of them, the 3rd one being the only one solved in an unworthy way, I've had a lot of fun!
The first one. I remember this happening. A work colleague told us he texted his friend at a minute pat 12 on January 1st telling his friend that next year he would be 50 next year! Obviously his friend had his 48th birthday the day before!
Number one could have been even better by double layering the riddle.
Consider all the stuff you said, but in addition consider how the phrasing in the original isn't "I will turn 16", it's "I will have my 16th birthday". Well when you turn 16, that means you have 'had' 17 total birthdays including the day you were born.
So the riddle could have been pushed even one year further while still being a perfectly logical statement "Two days ago I was 13 years old, but next year I will have my 17th birthday.""
I thought the same thing! 🧠
_Birthday_ means the *anniversary* of the day on which you were born. The very day on which you were born is not an anniversary of that birth, hence it's not a "birthday".
The literal day you were born is your birthdate, not your first birthday.
Puzzle 4 : Like you, Presh, I got this one wrong.
I wish you had told us exactly how you went wrong, just in case my mistake was the same.
Sometimes we learn more from our mistakes than we do from understanding the correct solution.
Oh yes! One more thing.......
"HAPPY NEW YEAR!" to you and all your readers.🥳🥳🥳
#4 Mary's age today represents 2X what John's age when Mary was X. Age difference is X which makes John's age today 3X. When Mary is 3X John will be 4X which sums to 7X. For 7X =63 the age difference is 9. Today Mary is 2X or 18 and John 3X or 27
Bro just has 2 extra birthdays in the same year
14:08 "If the house number was anything besides 13, the census taker would be able to solve for the ages."
Not true. If the house number was 21, the kids' ages could have been {1 , 2 , 18} , but also {½ , 4½ , 16} ; which means the census taker wouldn't have been able to solve for the ages at that point.
Good point. I wonder if there are any other solutions with fractional ages?
@mb-3faze I also got
{1½ , 1½, 16} , in case the house number is 19
but this is then not a doubled house number.
Other fractional triples with product 36 and that lead to an integer sum:
{37.5 , 3.2 , 0.3}, with house number 41
{⅓ , ⅔ , 162} (with house number 163) but this is not realistic
{ 98/5 , 50/7 , 9/35 } (house number 27)
{ 196/5 , 75/7 , 3/35 } (house number 50)
{a*p²/q , b*q²/p , c/(pq)}
with positive integers p and q such that {p, q, 6} are pair-wise co-prime*, and a, b, c are positive integers such that abc = 36 and
(a*p³ + b*q³ + c) is a multiple of pq .
There are probably many more.
[*] By saying "{p, q, 6} are pair-wise co-prime", I mean:
gcd(p,q) = 1 & gcd(p,6) = 1 & gcd(q,6) = 1
@mb-3faze Other fractional triples include:
{1½ , 1½, 16} with house number 19
{37.5 , 3.2 , 0.3} with house number 41
{⅓ , ⅔ , 162} (house number 163) but this is not realistic
{ 98/5 , 50/7 , 9/35 } (house number 27)
{ 196/5 , 75/7 , 3/35 } (house number 50)
but these are then not doubled house numbers.
@@mb-3faze I posted two replies with other (positive) fractional triples whose product is 36 and whose sum is an integer, but for some reason youtube hides these replies (unless the comment section is sorted in "Newest first" order instead of "Most popular first" order).
Three of the triples are {1½ , 1½, 16} , {37.5 , 3.2 , 0.3} and { 98/5 , 50/7 , 9/35 } .
Two useful words that have fallen out of common usage: ereyesterday for the day before yesterday and overmorrow for the day after tomorrow.
You failed to mention the twins in the setup of Puzzle 5
Puzzle 1: Yesterday, easy, since the puzzle doesn't specify the calendar date as a requirement.
Alternative answer to #1: They're living on a moon colony.
And they are really, really smart compared to the other 15-year-olds there.
If you allow that their are colonies on the moon, which is already fiction, then how does this affect their age anyway? And how could them being "smart" affect their age?
The moon rotates in the same time it takes to spin, which means a moon day is equivalent to a moon year (and since both are about 28 days, a 15 year old would instead only be about a 14 month old in earth-time development
@@Altemeous If a human is on the moon they are still counting time the exact same as on earth. A year is the time it takes for EARTH to travel around the sun. Other planets or moons dont have their own unique years. We measure their individual orbits in our own earth years.
the father's age problem can be solved analytically by comparing the coeficients of -9 and 10(x-2y)+y-2x. the statement -9 = 10(x-2y)+y-2x follows directly from 10x-10(2y)-2x+18 - 9 = 0, which follows directly from 10x+y-9 = 2(10y+x-9). we conclude from the statement -9 = 10(x-2y)+y-2x that x - 2y = 0 and y - 2x = 9. this implies that y - 2(2y) = - 9, which implies -3y=-9. hence y = 3. x=2y implies that x = 6. hence the father's age is 63
For the last one i solved it by the guy saying his eldest son was excited to have siblings, meaning plural, meaning twins since he knew he would get more then 1
2:14 sunday
The first problem makes me think the person’s birthday takes place on Mercury.
Me, 10: second one is easy!
Him (idk): **EXPLAAAIIINNSSSSSSSSSSSSSSSSSSS**
12:25? Why did you just assume that there were twins? I know I took the same assumption, but you didn't specify this assumption before casually mentioning it in an other assumption
Because in cases of no twins the census maker would be able to know the ages of the mathematician children. The two cases of the twin siblings are the two only cases where the Third hint that the mathematician gave to the census maker is necessary.
You can't have 2 birthdays in the same year. If the current year is year 1 then you need to turn 16 in year 2, turn 15 in year 1, and turn 14 in year 0.
If you were 13 years old 2 days ago and you're in year 1 that means that the only day that your birthday could be is December 31. If it's before 31.12 then you'd currently be in year 0. If it was after, then your 14th birthday would be in year 1.
Puzzle one. I’m lost. How does one day equal a year?
I laughed at the first one, my birthday is Jan 1st and my brother's is Dec 30th.
I'm not going to exactly remember the setup or the specific numbers in the original but the best age riddle I ever heard was something crazy like:
Jimmy's age in 5 years will be the same as if you took his mother's age, subtracted 3, halved it, subtracted 3 again, and halved it again. If Jimmy's mother is 26, where is the father?
In a way, this one is solvable 😂 I gave part of it away by implying that the numbers matter. No misdirection there 🧠
Spoiler - answer ahead:
Most sane people will look at it and dismiss it as nonsense, but then you start to work it out and you find that Jimmy's age works out to be -¾.
Huh. Well then something must have gone wrong with the writing of the riddle. According to the math, Jimmy won't even be born for another 9 mon- 😮
@@DrEcho -- Exactly. Jimmy was either just conceived or hasn't been conceived quite yet. In the latter case, there is as of yet no father. (I'm not going to add the complication of calculating weeks instead of months for a full-term baby Jimmy.)
@@DrEcho definitely ambiguous, but if you make the blatant assumption on the average time of pregnancy being exactly 9 months then I guess you can make a joke related to how they are having sexual intercourse right at the very moment Jimmy goes into existence, im gonna guess that's the idea
@@DrEcho I thought the answer would be something like the mother being 13 yo when conceiving Jimmy, and therefore the father would be on jail. But the actual solution is more hilarious 🤣
@randomgamer-te8op most definitely there are a few big assumptions, not least the exact gestational age of Jimmy at birth. I'm guessing that the idea of the joke is once you infer and take for given that there even is a somewhat cogent solution, you can answer with a bit of certainty after you overcome the math barrier. Akin to puzzles like last problem in this video; only when you're given the instance that you have enough information to solve it, can you actually solve it.
But, yes, you are correct that the alternative 'correct' answer is "the father could be literally anywhere, as we do not have the information required to make such a deduction".
Solved all but the last.
Puzzel 4 totally fried my brain.