The choices made it much easier. I estimated, that average of min. and max. is approx 50 and there is approx. 100 numbers total. Therefore 50x100 is 5000 and D is close, as the others are too far from the quess. But that is just an engineering approach to fast check some calculations, not a real math. I like you cover many different math topics and it returns me in my school years and discovering lost knowledge, but in a very different light. That is refreshing.
@@TonyWhitley no, the sort of engineering maths that ensures you didn't stuff up your working. Is your properly worked out answer close to the guestimate?
@@TonyWhitley Yours is the sort of comment made by someone who has never used the simple "Rule of thumb" to check a more complex, and time consuming, calculation.
This is exactly how I solved the question in 2 minutes. You are not being asked to solve the exact answer, just pick one of four answers. And 5050 was the only answer that was close. A good engineer will always estimate a solution to make sure the exact answer they calculate makes sense.
Admittedly I would’ve choked on that question. But, in guessing mode I would’ve gone with 5050 just because I doubted the least significant digit would be 3,4, or 8. I like this channel. Kind of a wake up call reminding me of the math many of us have forgotten by becoming software junkies.
Well imagine you’re in the hot seat … I went 5050 just because the least significant digit is 0. The 3,4, and 8 integer was the real give away when you only have 20 sec to respond.
Im a civil engineer and I have to do quick and dirty approximations on the fly all the time, until I get back to my office to check properly. For this: 1 to 100 roughly averages to 50. 50 x 100 = 5000 approx so 5050 is the closest answer.
Another way to calculate: We see that the simultaneous addition of the small numbers from 1 to 49 with the large numbers from 99 to 51 gives a constant figure of 100 * 49 = 4,900. We add (50 +100) + 4,900 We will therefore have the total of 5,050
I started by adding the largest numbers first knowing it be near 1000 just for the 100+99+98… and again near another 1000 for the 89+88+87… so I knew I was way over 3000.
I first added the numbers 1 to 10 in my head. 45. Repeat that ten times gets to 450. Adding 10, 20, 30 …. 90 is the same process multiplied by 10. 4,500. 450+4,500+100 comes to 5,050. I love your problems and clear explanations. I have not taken any math since high school over 55 years ago. Thank you for helping keep my brain exercised.
I looked at the problem and did it the same way she explained because that felt natural to me but it's amazing getting to see other people have completely different ways of thinking about the problem that get to the same solution.😀
Like Mr. Gauss, I also “derived” the same solution in elementary school. My formula, however, was generally valid for any length string of consecutive numbers. Since I am sure that I am not the smartest kid in the room, I also believe that tens of thousands of other school kids have figured it out.
The formula was around since the ancient Greeks. I have no idea why this story of the whole world not knowing this until some kid worked it out for them keeps being retold. It's like really believing an apple fell on Newton's head.
Gauss is considered as a greatest mathematician from Germany. Wonderful,he solved it during his school days. Recently,I solved it after Gauss as 1+2+..... ..50 100+ .. +51 ----------------------- 101+101+. . +101=50×101=50(100+1)=5050.
My grade 6 teacher gave us the exact same problem to solve. He stopped us after a few minutes and told us the same story of Gauss. I recognized the problem/solution right away and solve it the same way in a few minutes. 😀
I remember learning something like this in algebra class. The teacher, Mister Olson, called it the "Blackjack formula" or something. Eks squared plus Eks, divided by two. So with that, it took me like five seconds to some up with, well, what's half of ten thousand one hundred? 5050. Answer is D. Sounds like Mister Olson could have made a lot more money going on that TV quiz show than teaching us algebra, or playing blackjack.
That was really cool! I took a different approach: I added the numbers 1 thru 10 to get 55. I then inferred that each group of 10 would end in 5 as well - thus, the sum of ten groups of ten would end in zero (five times ten). Since option D was the only number to end in zero, I knew it had to be option D
Generally, sum integers from a to b. Make pairs. The sum is the total of each pair times the number of pairs. There could be a number in the middle of the sequence left unpaired, this can be considered to be half a pair. So ... (b+a)(b-a+1)/2. For the example in the video: (100+1)(100-1+1)/2 101*100/2=5050 If we were adding up 100 to 200, then it would be: (200+100)(200-100+1)/2 300*101/2 (50 pairs adding to 300, plus a 150 left over) 15150
I find it impossible to believe no mathematicians before could figure out a simple formula for adding sequential numbers until some kid came along. In reality the older Gauss liked to tell people this story because he was proud of what he did at such a young age. Not that he was the world's first to have discovered it.
Whacky, I came up with that exact formula as a kid when asked to add 1-10, then try 1-100. (Decided to come up with a formula for the longer one). Btw 1-10 made it an obvious formula (11x5) or 55. First I came up with 2 formulas, one that ended in an odd high number, then one that ended in an even higher number. Then figured out a formula that would work for either X(X+1)/2 If anyone is curious: Formula for a series that ends in an EVEN number where N denotes the high number. Formula is = N(N/2) Formula for a series that ends on an ODD number where N is the high number. Formula is N (N/2 +.5) Formula for either N(N+1)/2
I did averages. Average 11-20 is 15, 21-30 is 25, 31- 40 is 35….. then 15x 10, 25x10, 35x10…. That lot came to 4950. Then add 1-10 and that gives you an inaccuracy of 50 ish.
For a moment, I thought you were about to present a solution to one of the Millennium Prize Problems! For those who are curious, these are a set of 7 unsolved mathematical problems. So far, only one of these problems has been solved (the Poincare Conjecture), but the mathematician who solved it declined a $1 million prize!
Solution: Answer D is correct. I don't even need to calculate, I remember it. But calculations: 1 + 100 = 101 2 + 99 = 101 3 + 98 = 101 ... 49 + 52 = 101 50 + 51 = 101 So you end up with 50 pairs of 101 or 50 * 101 = 5050 Even easier to use Gauß: x = n(n + 1) / 2 = 100 * 101 / 2 = 50 * 101 = 5050
The multiple choice answers made it too easy. I knew instantly that it would end in a 5 or a 0, and there's only one answer that works. Would have definitely had a tougher time if I actually had to calculate the answer :P
Without any calculation I know the answer is D. This question is familiar to me: break the row in 2, from 1-50 and 51-100, add the lowest with the highest so 1+100, 2+99, 3+98 = 101 until 50+51 so 50 . 101 = 5050. Nice story about Carl Frederick Gauss.
Got the answer in about 3 seconds. I could see the midpoint was roundabout the 50 mark. There are 100 numbers, therefore 100 x 50 = 5,000. That rules out answers A, B and C, leaving D.
il suffit de découper la série en 2 et de les additionner par paire : 1+99, 2+98, 3+97, etc... il restera la valeur 50 seule, donc 50 couples * 100 + 50 = 5050
Or you can look at this and realize that it's an arithmetic series. The general formula is 1st term plus last term which is 1 and 100 respectively. Then divide by two to get 50.5,and then multiply by the number of terms which is 100 to get 5050. The formula that is used in this video only works if the 1st term is 1. Same idea basically.
I reasoned this out before I started the video. Just add the lowest and highest number and multiply by the number of pairs. Because the first number is odd and the last number is even, the number of pairs will be an integer (in this case, 50). But it took a couple of minutes for me to recognize this so I probably would have had to burn a lifeline to gain the extra time needed. Or maybe I could have stalled long enough, idk. Adding integers in a series becomes a bit more interesting if you aren't starting from 1, though.
Very easy question I haven't bothered watching but calculated 5,050 in about 5 seconds.Edit; just read some comments, so many different ways Edit; I just read some comments, so many different ways suggested. I did 1+100=101, 101x50 pairs-5,050. Easy to do it in your head.
(101)(50)=5050 simple. According to the story, Gauss used this technique to solve this problem during elementary school. His teacher was trying to keep the students busy for a while and asked the class to find the sum of the first 100 positive integers. Guess quickly figured out that pairing off the numbers sequentially (1 + 100, 2 + 99, …) gave 50 pairs of 101. Thus (101)(50) yields the result. I will assume that your video demonstrates this technique and wins the $32,000.
In the original telling of the story, Gauss' technique was not mentioned and there are several ways to do it, the motivation for the assignment was not mentioned, and which numbers had been added was also not stated. The only thing that was said was that Gauss finished an arithmetic sum quickly. All other details were made up by various different authors decades or centuries later. Often conflicting with other accounts.
I have a set of formulas that i derived back in 1979 that I called my Prince "Dixie cup" Formulas. They are the formulas needed for determining the flat shape (diameter and pie cut out of it) for making a cone of a given diameter and angle of the cone. Let me know how to get it to you if you are interested in seeing it and/or making a video of it. At that time, and maybe still, I had never found equations for solving this problem.
I solved this problem by noticing that the last 50 numbers (50-100) have 50 in them and I multiplied 50 times 50 to get 2500. I then did not see that the answer C was 3, 333 and so I guessed correctly that the only larger number than 2500 available (D; 5, 500) must be the correct answer. Nothing hits the spot quite like doing the calculations *so* wrong that you get the right answer anyway... 😁
Hi Queeny, I like the ease and clean simplicity of this alterative. Just add 1 to 99 = 100, 2 to 98 = 100, etc. This operation renders us 49 pairs equaling 100 or 4900. Then all that's left is to account for the unpaired 50 and the unpaired 100 which sum to 150. So now we sum 4900 and 150 to arrive at 5050. QED! 😀
Had the problem been to add 100 with itself 100 times, the answer would have been 10'000. The terms of the problem slowly increase from almost nothing to 100, so you get sort of a triangle with the same base and height as the square (100 x 100 = 100² = 10'000). The triangle covers half the square, so it must have half the area, or half of 10'000. The answer should be close to half of 10'000, or 5'000. Only D is even close to that.
said 5050 even before I checked it with quick program in c. 100 x 100 would be 10000.. so intuitively its about half. Also if we just do very rough sums going form 100 down.. first 10 is about 1000.. first 20 about 2000.. first 30, closer to 40 about 3000., so its more than that.. leaves 5050
wow i got to the same answer in a weird way all in my head. i framed the problem as "half the numbers are above 50, and half below". the number 100 added 100 times (100x100=10,000). Then the number 1 added 100 times (100). Divide by 2.
I have a better option. Keep 100 as it is. Now add 1+99 which equals 100 and do it till 49+51. So we get 49 sums of 100. So we have now 49*100=4,900. Now add 100 and 50. Which makes 5,050. We can calculate this is few seconds. Hope this is also correct.
She did this the harder way. Add 1 and 99, 2 and 98,... each total is 100 all the way up to 49 and 51. Leaving you 50 and 100 never added. 4900 plus 150. Way simplar than 101.
I think I did it an easier way using an average: highest number is 100 x 100 = 10,000 lowest number is 1 x 100 = 100 Average of those two numbers = (10,000 + 100)/2 = 5050
Hi Queeny; I chose the right answer but for the wrong reason. I thought that we are adding 100 integers together, and those integers have an average value of 50. Therefore 50 * 100 = 5000. Why is this wrong ?
[edit - realised I forgot to say why the average was wrong] It's wrong because the average would be 50.5, not 50. (100+1)/2=50.5 (99+2)/2=50.5 ... (51+50)/2=50.5 Average = 50.5 If you were adding up 1 to 99, then the average would be 50 (49 pairs adding to 100 and a 50 left unpaired), and 99*50=4950. (Adding the 100 onto this gives the 5050, which is the sum of 1 to 100.) You've just made a mistake with working out the average, that's all. Let's work out the sum of the numbers from 4 to 8 (easy enough to do by adding each number in turn to check - the answer should be 30). Using your method: Average = 6 [(4+8)/2, this only works because it's a complete sequence of integers, we can't generally calculate the average of a set of numbers by adding the max and min and dividing by 2]. Number = 5 [8-4+1, or 8-(4-1) if you prefer, again because it's a complete sequence]. Sum = 6*5 = 30. Using the video method: Paris add to 12 [8+4, 7+5]. 6 is left over unpaired (didn't need to be mentioned in the video as there wasn't one left unpaired). Number of pairs = 2. 2*12+6=30. Using a general method based on the pairing: Sum integers from a to b. Total of each pair times the number of pairs (including counting an unpaired number left over in the middle of the sequence as half of a pair). (b+a)*(b-a+1)/2. For our example. (8+4)*(8-4+1)/2 12*5/2 = 30
n(n+1)/2= 100*101/2=5050. Die Geschichte des kleinen Gauß in der Schule. War ja als Strafe-Aufgabe gedacht. Den Gewinn dann bitte auf das Konto überweisen :D
The choices made it much easier. I estimated, that average of min. and max. is approx 50 and there is approx. 100 numbers total. Therefore 50x100 is 5000 and D is close, as the others are too far from the quess. But that is just an engineering approach to fast check some calculations, not a real math.
I like you cover many different math topics and it returns me in my school years and discovering lost knowledge, but in a very different light. That is refreshing.
The sort of engineering maths that resulted in the Mars Climate Orbiter crashing into Mars😛
@@TonyWhitley no, the sort of engineering maths that ensures you didn't stuff up your working. Is your properly worked out answer close to the guestimate?
@@TonyWhitley Yours is the sort of comment made by someone who has never used the simple "Rule of thumb" to check a more complex, and time consuming, calculation.
Yours is the sort of comment made by someone who has no sense of humour.
This is exactly how I solved the question in 2 minutes. You are not being asked to solve the exact answer, just pick one of four answers. And 5050 was the only answer that was close. A good engineer will always estimate a solution to make sure the exact answer they calculate makes sense.
A classic problem solved in a very comprehensive and clear way again, top Suzanna!
Admittedly I would’ve choked on that question. But, in guessing mode I would’ve gone with 5050 just because I doubted the least significant digit would be 3,4, or 8.
I like this channel. Kind of a wake up call reminding me of the math many of us have forgotten by becoming software junkies.
As a multiple choice it is easy, average value of the largest and smallest is 50.5 and there are 100 of them
That’s how I did it too. Average is just over 50 so the sum is just over 5000, and there was only 1 answer like that.
That's also how I did it.
Right!? 😂
Well imagine you’re in the hot seat …
I went 5050 just because the least significant digit is 0. The 3,4, and 8 integer was the real give away when you only have 20 sec to respond.
Im a civil engineer and I have to do quick and dirty approximations on the fly all the time, until I get back to my office to check properly. For this: 1 to 100 roughly averages to 50. 50 x 100 = 5000 approx so 5050 is the closest answer.
my way was similar. add 100, 100 times. add 1, 100 times. divide by 2 for average.
5050 is not the closest. It's the exact
Another way to calculate: We see that the simultaneous addition of the small numbers from 1 to 49 with the large numbers from 99 to 51 gives a constant figure of 100 * 49 = 4,900. We add (50 +100) + 4,900 We will therefore have the total of 5,050
I started by adding the largest numbers first knowing it be near 1000 just for the 100+99+98… and again near another 1000 for the 89+88+87… so I knew I was way over 3000.
Math is not one of my best competences but you have a way of simplifying things so that I do understand it. Thank you. 🙂
I first added the numbers 1 to 10 in my head. 45. Repeat that ten times gets to 450. Adding 10, 20, 30 …. 90 is the same process multiplied by 10. 4,500. 450+4,500+100 comes to 5,050.
I love your problems and clear explanations. I have not taken any math since high school over 55 years ago. Thank you for helping keep my brain exercised.
I looked at the problem and did it the same way she explained because that felt natural to me but it's amazing getting to see other people have completely different ways of thinking about the problem that get to the same solution.😀
Correction: The first step is 1 to 9. I wrote it wrong but did it right.
Very clever, sir. I like how many math problems can be solved using several different methods.
@@jackmachen7390 1+2+3+4+5+6+7+8+9+10
(10+1)+(9+2)+(8+3)+(7+4)+(6+5)=11+11+11+11+11=5×11=55.
I approached this problem in the same way.
great story telling, you're making me fall in love with math
Like Mr. Gauss, I also “derived” the same solution in elementary school. My formula, however, was generally valid for any length string of consecutive numbers. Since I am sure that I am not the smartest kid in the room, I also believe that tens of thousands of other school kids have figured it out.
The formula was around since the ancient Greeks. I have no idea why this story of the whole world not knowing this until some kid worked it out for them keeps being retold. It's like really believing an apple fell on Newton's head.
Gauss is considered as a greatest mathematician from Germany. Wonderful,he solved it during his school days. Recently,I solved it after Gauss as 1+2+..... ..50
100+ .. +51
-----------------------
101+101+. . +101=50×101=50(100+1)=5050.
I think that is the easiest method if you are not familiar with the formula.
Our math teacher told us the exact story with Gauss, when I was in school, so when you presented the problem, I knew.
My grade 6 teacher gave us the exact same problem to solve. He stopped us after a few minutes and told us the same story of Gauss. I recognized the problem/solution right away and solve it the same way in a few minutes. 😀
You are the perfect math teacher we all wish we had. And such a beautiful voice. I am in love.
Get a grip!
I remember learning something like this in algebra class. The teacher, Mister Olson, called it the "Blackjack formula" or something. Eks squared plus Eks, divided by two. So with that, it took me like five seconds to some up with, well, what's half of ten thousand one hundred? 5050. Answer is D. Sounds like Mister Olson could have made a lot more money going on that TV quiz show than teaching us algebra, or playing blackjack.
Took me 2 minutes in my head to get to the same conclusion using difference of squares, (50+1)(50-1)
how is that relevant?
That was really cool! I took a different approach: I added the numbers 1 thru 10 to get 55. I then inferred that each group of 10 would end in 5 as well - thus, the sum of ten groups of ten would end in zero (five times ten). Since option D was the only number to end in zero, I knew it had to be option D
Any number times ten will end in a zero. The only option had to be D.
I love this channel!
My favorite math teacher of all time (Professor Haenisch, Trenton State College) told us that story about Gauss. He adored Gauss!
Generally, sum integers from a to b.
Make pairs. The sum is the total of each pair times the number of pairs. There could be a number in the middle of the sequence left unpaired, this can be considered to be half a pair. So ...
(b+a)(b-a+1)/2.
For the example in the video:
(100+1)(100-1+1)/2
101*100/2=5050
If we were adding up 100 to 200, then it would be:
(200+100)(200-100+1)/2
300*101/2
(50 pairs adding to 300, plus a 150 left over)
15150
I find it impossible to believe no mathematicians before could figure out a simple formula for adding sequential numbers until some kid came along.
In reality the older Gauss liked to tell people this story because he was proud of what he did at such a young age. Not that he was the world's first to have discovered it.
Thank you, Susanne. ❤❤❤❤
Whacky, I came up with that exact formula as a kid when asked to add 1-10, then try 1-100. (Decided to come up with a formula for the longer one). Btw 1-10 made it an obvious formula (11x5) or 55.
First I came up with 2 formulas, one that ended in an odd high number, then one that ended in an even higher number. Then figured out a formula that would work for either X(X+1)/2
If anyone is curious:
Formula for a series that ends in an EVEN number where N denotes the high number.
Formula is = N(N/2)
Formula for a series that ends on an ODD number where N is the high number.
Formula is N (N/2 +.5)
Formula for either N(N+1)/2
Knowing that 1+99=100 and multiplying this by 50 would be all the contestant needs to answer with confidence.
This formula only works if your series begins at 1. I prefer to use first# + last# x half the number of numbers in the series.
I don't know how many times I looped 1:03 just to hear you pronounce Carl Friedrich Gauss... over and over and over! ;)
Haha, I can’t help the German in me 😅
@MathQueenSusanne lovely!❤
Simple but at the same time brilliant
I did averages. Average 11-20 is 15, 21-30 is 25, 31- 40 is 35….. then 15x 10, 25x10, 35x10…. That lot came to 4950. Then add 1-10 and that gives you an inaccuracy of 50 ish.
I just take the average of the lowest and highest, 50.5, and multiply by 100. Basically the same method at heart.
Regis would suggest A 50 50 split! Great explanation. Keep smiling............
For a moment, I thought you were about to present a solution to one of the Millennium Prize Problems! For those who are curious, these are a set of 7 unsolved mathematical problems. So far, only one of these problems has been solved (the Poincare Conjecture), but the mathematician who solved it declined a $1 million prize!
I love your videos. May the Lord continue to bless your life.. 🙏🙏☀☀
I just guessed 5050 immediately, not because I know math, but because 50 was the halfway point between 1 and 100.
Solution:
Answer D is correct. I don't even need to calculate, I remember it.
But calculations:
1 + 100 = 101
2 + 99 = 101
3 + 98 = 101
...
49 + 52 = 101
50 + 51 = 101
So you end up with 50 pairs of 101 or 50 * 101 = 5050
Even easier to use Gauß:
x = n(n + 1) / 2 = 100 * 101 / 2 = 50 * 101 = 5050
The multiple choice answers made it too easy. I knew instantly that it would end in a 5 or a 0, and there's only one answer that works. Would have definitely had a tougher time if I actually had to calculate the answer :P
Kleine Gauss awww! 🤍
Without any calculation I know the answer is D. This question is familiar to me: break the row in 2, from 1-50 and 51-100, add the lowest with the highest so 1+100, 2+99, 3+98 = 101 until 50+51 so 50 . 101 = 5050. Nice story about Carl Frederick Gauss.
Got the answer in about 3 seconds. I could see the midpoint was roundabout the 50 mark. There are 100 numbers, therefore 100 x 50 = 5,000. That rules out answers A, B and C, leaving D.
il suffit de découper la série en 2 et de les additionner par paire : 1+99, 2+98, 3+97, etc... il restera la valeur 50 seule, donc 50 couples * 100 + 50 = 5050
Or you can look at this and realize that it's an arithmetic series. The general formula is 1st term plus last term which is 1 and 100 respectively. Then divide by two to get 50.5,and then multiply by the number of terms which is 100 to get 5050. The formula that is used in this video only works if the 1st term is 1. Same idea basically.
I reasoned this out before I started the video. Just add the lowest and highest number and multiply by the number of pairs. Because the first number is odd and the last number is even, the number of pairs will be an integer (in this case, 50). But it took a couple of minutes for me to recognize this so I probably would have had to burn a lifeline to gain the extra time needed. Or maybe I could have stalled long enough, idk.
Adding integers in a series becomes a bit more interesting if you aren't starting from 1, though.
All the brains and all the beauty ❤❤❤
Have a good, numerically enjoyable, weekend, Ms. Susanne!
Thank you! You too!
finally, something i can understand!
Add the smallest to the biggest = 101
Divide the amount of numbers by 2 =50
Multiply the 2 results = 5050
Oddly, I still remember that formula from back in the day. I would have won that question, no problem.
Excellent
I learned that, from a math course on Brilliant. It's cool!!
Very easy question I haven't bothered watching but calculated 5,050 in about 5 seconds.Edit; just read some comments, so many different ways
Edit; I just read some comments, so many different ways suggested. I did 1+100=101, 101x50 pairs-5,050. Easy to do it in your head.
If you add the numbers from 1 to 100, the result is 5,050 final answer.
I won and donate it to your channel lol
Nice puzzle!
It being styled as a WWTBAM question is a big hint 🙂
(101)(50)=5050 simple. According to the story, Gauss used this technique to solve this problem during elementary school. His teacher was trying to keep the students busy for a while and asked the class to find the sum of the first 100 positive integers. Guess quickly figured out that pairing off the numbers sequentially (1 + 100, 2 + 99, …) gave 50 pairs of 101. Thus (101)(50) yields the result. I will assume that your video demonstrates this technique and wins the $32,000.
In the original telling of the story, Gauss' technique was not mentioned and there are several ways to do it, the motivation for the assignment was not mentioned, and which numbers had been added was also not stated. The only thing that was said was that Gauss finished an arithmetic sum quickly. All other details were made up by various different authors decades or centuries later. Often conflicting with other accounts.
I have a set of formulas that i derived back in 1979 that I called my Prince "Dixie cup" Formulas. They are the formulas needed for determining the flat shape (diameter and pie cut out of it) for making a cone of a given diameter and angle of the cone. Let me know how to get it to you if you are interested in seeing it and/or making a video of it. At that time, and maybe still, I had never found equations for solving this problem.
I solved this problem by noticing that the last 50 numbers (50-100) have 50 in them and I multiplied 50 times 50 to get 2500. I then did not see that the answer C was 3, 333 and so I guessed correctly that the only larger number than 2500 available (D; 5, 500) must be the correct answer. Nothing hits the spot quite like doing the calculations *so* wrong that you get the right answer anyway... 😁
I did it by ( 1+2+3 ...10 ) =55 : ( 11+12+13...20 =155 ) : (21+22+23 ....30 = 255) ... : : ... ( 55 +155+255+355 ...955 = 5050 ) : : 5050= Ans
Hi Queeny, I like the ease and clean simplicity of this alterative. Just add 1 to 99 = 100, 2 to 98 = 100, etc. This operation renders us 49 pairs equaling 100 or 4900. Then all that's left is to account for the unpaired 50 and the unpaired 100 which sum to 150. So now we sum 4900 and 150 to arrive at 5050. QED! 😀
Had the problem been to add 100 with itself 100 times, the answer would have been 10'000. The terms of the problem slowly increase from almost nothing to 100, so you get sort of a triangle with the same base and height as the square (100 x 100 = 100² = 10'000). The triangle covers half the square, so it must have half the area, or half of 10'000.
The answer should be close to half of 10'000, or 5'000. Only D is even close to that.
said 5050 even before I checked it with quick program in c. 100 x 100 would be 10000.. so intuitively its about half.
Also if we just do very rough sums going form 100 down.. first 10 is about 1000.. first 20 about 2000.. first 30, closer to 40 about 3000., so its more than that.. leaves 5050
Gefällt mir auf englisch genau so gut wie auf deutsch. 😊
That is great
wow i got to the same answer in a weird way all in my head. i framed the problem as "half the numbers are above 50, and half below". the number 100 added 100 times (100x100=10,000). Then the number 1 added 100 times (100). Divide by 2.
I have a better option.
Keep 100 as it is.
Now add 1+99 which equals 100 and do it till 49+51. So we get 49 sums of 100.
So we have now 49*100=4,900. Now add 100 and 50. Which makes 5,050. We can calculate this is few seconds. Hope this is also correct.
1/2n(n+1). 50 * 101 = 5050
The $ goes before the number. $32,000
Besides 100, every number has pair to make 100. That's 50 one hundreds. So being multiple choice it is easy to get the answer
Does the US version of Who Wants to be Millionaire have a time limit as the British version doesn't, so this seems a strange question to ask?
😈 Please, never add the numbers from 1 to 36. Do not try. No, no, no.
\m/
🤘🤘🤘
Or 1/2 N(N+1). You are so cool btw
Thanks!
Susanne, can you make video about "e" number? I think, it would be interesting.
Even if you don't know Σ1..n=(n²+n)÷2, a reasonably curious person could easily see that A and B are ‘too convenient’, and C is unlikely.
(N (N+1 ))/2 is the formula. (100 * 101) / 2 = 5050
Cool, if you don't remember sum of arithmetic sequence formula
WWGD? Famous story. 😊
Which Carl Gauss did in his head aged 7.
Maybe. Or 9 or 11. Or he worked it out on a board. There are hundreds of accounts of this story with slight variations to them all.
2 seconds, final answer, D)
I'm thinking about a formula for starting with a number other than 1.
x + (x+1) + (x+2) +......+ (n-1) + n
adding the numbers between any two numbers
The square root of the current year, 2025 is 45 and also the sum of the numbers 1 through 9 as any Killer Sudoku solver should know.
She did this the harder way. Add 1 and 99, 2 and 98,... each total is 100 all the way up to 49 and 51. Leaving you 50 and 100 never added. 4900 plus 150. Way simplar than 101.
Its quite close to the area of a triangle
Very nice! But it's a repeat for me.
Great video the thing that bothers me is all the answers list have a decimal point instead of a comma.
What about all the numbers BELOW 50?
I grew up with the formula: sigma(1 to n)=(n*(n+1))/2
I knew the answer as I knew the story.
Вместо Moonsun я слушаю математику😮
I think I did it an easier way using an average:
highest number is 100 x 100 = 10,000
lowest number is 1 x 100 = 100
Average of those two numbers = (10,000 + 100)/2
= 5050
That "divide by 2" in Gauss' formula IS an average.
@@Lynn-u7l but it seemed more complex. or maybe im just simple.
N(n+1)/2
Hi Queeny; I chose the right answer but for the wrong reason. I thought that we are adding 100 integers together, and those integers have an average value of 50. Therefore 50 * 100 = 5000. Why is this wrong ?
The premise misleads. The average won't be a whole number integer of 50, it'll be a decimal of 50.5.
[edit - realised I forgot to say why the average was wrong]
It's wrong because the average would be 50.5, not 50.
(100+1)/2=50.5
(99+2)/2=50.5
...
(51+50)/2=50.5
Average = 50.5
If you were adding up 1 to 99, then the average would be 50 (49 pairs adding to 100 and a 50 left unpaired), and 99*50=4950. (Adding the 100 onto this gives the 5050, which is the sum of 1 to 100.)
You've just made a mistake with working out the average, that's all.
Let's work out the sum of the numbers from 4 to 8 (easy enough to do by adding each number in turn to check - the answer should be 30).
Using your method:
Average = 6 [(4+8)/2, this only works because it's a complete sequence of integers, we can't generally calculate the average of a set of numbers by adding the max and min and dividing by 2]. Number = 5 [8-4+1, or 8-(4-1) if you prefer, again because it's a complete sequence].
Sum = 6*5 = 30.
Using the video method:
Paris add to 12 [8+4, 7+5]. 6 is left over unpaired (didn't need to be mentioned in the video as there wasn't one left unpaired). Number of pairs = 2.
2*12+6=30.
Using a general method based on the pairing:
Sum integers from a to b.
Total of each pair times the number of pairs (including counting an unpaired number left over in the middle of the sequence as half of a pair).
(b+a)*(b-a+1)/2.
For our example.
(8+4)*(8-4+1)/2
12*5/2 = 30
1+99=100 2+98=100… There are 49 pairs of 100 which equals 4900. Then add 50 and 100 since they were not included in the 49 pairs.
Divide and conquer… 50 x 101
The right answer is D) 5.050
100 x (100 + 1) /2 =5,050
n(n+1)/2= 100*101/2=5050. Die Geschichte des kleinen Gauß in der Schule. War ja als Strafe-Aufgabe gedacht. Den Gewinn dann bitte auf das Konto überweisen :D
It is NOT as simple as that...still adding.....
n/2(1+100)=100/2(101)=5050
(1+100)/2*100
Eighth grade highschool knowledge.
50 pairs of 101 so 5050
Spoiler alert 😂😂😂