A legendary question with a surprising answer

Thanks to Jarek for the suggestion! What's a formula for the maximum product of a partition of a number? This was an Olympiad question and a Putnam problem! Special thanks this month to: Kyle, Lee Redden, Mike Robertson, Daniel Lewis. Thanks to all supporters on Patreon! / mindyourdecisions
0:00 Problem
1:11 Solution
6:40 Intuition
Math StackExchange
math.stackexchange.com/questi...
math.stackexchange.com/questi...
OEIS A000792
oeis.org/A000792
Putnam 1979, A1
prase.cz/kalva/putnam/psoln/p...
Maximum integer product
www.geeksforgeeks.org/breakin...
IMO 1976, B1
prase.cz/kalva/imo/isoln/isol...
IMO 1976, 4
artofproblemsolving.com/wiki/...
Product-maximising partitions
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The MaxProduct Partition
www.jstor.org/stable/23249541
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[@kevinmartin7760]

You don't want the number closest to e, you want the integer on either side of e that yields the maximum value for the function. These are not necessarily the same (the function may drop quickly above e and slowly below it) but in this case they are. The can be established by evaluatingt the function at both x=2 and x=3.

[@MasterHigure]

Depends on how you measure "closest". In this particular instance, we clearly see that 2 and 4 are equally close e, so 3 must be closer, as the distance function in question is strictly monotonic.

[@mchammer5026]

​@@MasterHigureop already established that in this particular case it works out correctly, but in the general case (for which the function doesn't need to be monotonic) you'd want to check both sides of e

[@lucianoneto4721]

​@@MasterHigure "we clearly see that 2 and 4 are equally close" would be the evaluation for x=2 and x=4 anyway, so the point op made still stands, you'd have to evaluate on 2 and 3 instead of assuming it

[@MasterHigure]

Yes, yes, obviously it needs to actually be checked to be rigorous. But if you're not occasionally allowed to intuit things in mathematics, you'll be stuck in the mud and get nowhere.

[@michaeledwards2251]

It doesn't state all the positive integers are the same, some can be 2 and some 3. Given 6 = 2+2+2 whose product is 8 or 3+3 whose product is 9, the problem reduces to how to maximise the number of 3s. This gives 3 power (1976 integer divide by 3, call it A). The remainder is 2. The result is 3 power A times 2. The reason why 3 is preferable is because base 3 requires the least number of states to represent positive integers.

[@MushookieMan]

Finally, engineers have been proven correct. Next please show that pi=4

[@finmat95]

pi = e = 3.

[@justpaulo]

This has indeed a practical application in an engineering problem in digital electronics: the scaling of a chain of inverters to drive a certain load with minimum propagation delay. Each inverter in the chain should be "e" times bigger than its antecessor. In practice 3 is usually the scaling factor chosen. In this application P would be the final load and S the propagation delay that you want to minimize given P.

[@alexanderhjertkvist7537]

⁠@@justpauloyou lost me at “this”😂

[@justpaulo]

@@alexanderhjertkvist7537 😂

[@ulmersapiens]

No engineer thinks that pi=4, because pi=3 to within 5%.

[@deathstroke7316]

*Maths problem has anti-derivatives, vector Laplacians, triple integrals,use of stokes' theorem, fourier transformations and divergence theorem all together* Presh: this problem was found in 3rd Grade math olympiad

[@Aim54Delta]

Now it makes sense... How I interpreted the question: "What is the largest product which can be achieved from numbers whose sum is 1976?" The answer is 2^988. 1976/2 is 988, so 988 2s exist in 1976, therefore the product of 988 2s recursively is 2^988. This guy went a completely different direction that was never intended for the test.

[@michaeledwards2251]

@@Aim54Delta The answer is 3 power (1976 integer divided by 3 ) times 2. (2 is the integer remainder from dividing 1976 by 3. ) Why 3, because 6 = 3 + 3 whose product is 9, or = 2 + 2 + 2 whose product is 8. The problem reduces to maximising the number of 3s. (It doesn't say the integers are all the same. ) It suspect the given solution to the answer was 2 power 988, 1976 / 2, but that is not the correct answer.

[@samiraperi467]

@@michaeledwards2251 You didn't watch the video, did you?

[@garyw.9628]

I guess you know that his answer is much higher ( as can be easily calculated on a calculator using logarithms )@@Aim54Delta

[@garyw.9628]

I guess you know that his answer is much higher ( as can be easily calculated on a calculator using logarithms )@@Aim54Delta

[@justpaulo]

I can relate this to a practical engineering problem in digital electronics: the scaling of a chain of inverters to drive a certain load. If you want minimize the propagation delay of such chain you should scale up by "e" consecutive inverters. In practice the inverters are usually scaled by a factor of 3. (in this case P would be the final load and S the total propagation delay... so instead of wanting to maximize P for a given S, we want no minimize S given a certain P, which is essentially the same)

[@romain.guillaume]

I solved it looking for the maximum integer image of the function ln(x)/x. The maximum of the function is e which is between 2 and 3, so we simply compare 2^3 and 3^2 and deduce that 3 is the optimum. Thus, it remains to breach down the case when the remainder mod3 is not 0 and don’t forget we cannot multiply by 1. We quickly conclude that you better complete with 2 (remainder 2) or 2*2 or 4 (same result for remainder 1). The rest is trivial and the results are 2*3^658 and 2*3^659

[@kcolloran]

This is a great problem. Solving the first problem is something I could work through with an older elementary or middle school student but yet understanding the why of it would be complicated for most high school students. Something with that range is so great as a creative puzzle for math students.

[@jeremiahlyleseditor437]

This was great. Might have been your best math theory problem ever.

[@tschantz]

I interpreted the question to be which TWO different positive integers whos sum is 1976 yields the highest product. Which would be 989 and 987. But…since we can use as many integers as we like, the answer is 658 threes and 1 two? Is that right? The sum of those 659 integers (threes and a two) adds up to 1979 and yields a product of (3^658)(2)=a 314 digit number that starts with 88,264,…

[@Abhishek-tq8hb]

I also did the same.

[@vestieee5098]

Very interesting method to solve this! Great job

[@kratosAk4700]

amazing......I figured out 3 was the key but didn't know why....the way you explained was amazing....I wish I had a math teacher like you when I was in my school.....

[@MichaelRothwell1]

Thank you for this wonderful problem and impeccable solution. I took the same route, both to solve the problem and identify what is special about the number 3 that appears in the solution.

[@leickrobinson5186]

At 0:50, I think you accidentally left out “and P is maximized”.

[@omarleonidez7200]

Thanks so much for the solution

[@cpsof]

Just for comparison: 2 * 3^658 ~= 10^314.25. Whereas 1976/e ~= 727 and (1926/727)^727 ~=10^315.70. So in the continuous case, the product is almost 30 times larger.

[@TimL_]

Brilliant problem! Best video in a while, well done.

[@lordluxembourg8777]

If we weren't limited to integers in this question then taking e^1976/e would yield a larger result, but since 3 is the closest integer to 3, that is of course the number we must use!

[@mathmannix]

Oh, I liked this one a lot. Not another five-minute puzzle I can easily solve on my own with basic algebra or trigonometry, but something really interesting. So I guess what we really want, if all values can be real numbers instead of merely integers, is to calculate 1976/e ~= 726.93..., and then say the product is e^726.93.

[@yurenchu]

The product cannot be e^(726.93...) though, because we can partition 1976 only into an _integer_ number of (positive) parts (so for example 725 parts, or 726 parts, or 727 parts, or 728 parts , or...). However, no matter how we partition 1976 into positive parts, the product cannot exceed e^(1976/e) .

[@ananyagupta1409]

There was a similar leetcode problem like this!

[@joelleet554]

tbh I thought 2^988 was so high of a number that I just stopped there and got proven wrong by several branches of mathematics working in conjunction The e outa nowhere reminds me of a dnd game from so long ago when the DM wanted a PC to choose two numbers between 1 and 4 (inclusive) so a d4 roll could determine his fate. If he rolled a number he chose it would be curtains for him. He picked e and pi.

[@leif1075]

Isn't that a valid answer too?

[@Crestalus]

My intuition said to spread them out as evenly as possible without them going to 1 (then the answer is just going to be 1) So, intuitively, I choose 2 as well 😭

[@hevado01]

@@Crestalusthought the same until I checked for 12: 3^4>2^6=4^3>6^2

[@yurenchu]

That was my first approach too. However, after that I immediately noticed that 2+2+2 = 3+3 , but 2*2*2 < 3*3 ; so dividing 1976 into 3s instead of 2s would give a greater result. (And dividing 1976 into 4s would give the same result as dividing into 2s, because 2+2 = 4 and 2*2 = 4 .)

[@JMurph2015]

​@@leif1075no because the answer using 3's is higher and therefore correct.

[@SuperPassek]

Nice analysis. At 10:30, though it is true that 3 is the closest number to e, considering 2 is in the opposite side of 3 from e, we can use that x = 2 is the same case as x = 4 and 4 is in the same side as 3 from e so 3 is definitely better than 4.

[@mrosskne]

Another way to convince yourself that 3 is always better than any larger number : First, as stated, 4n and (2)(2)n are equivalent. If you have a 5, that can be turned into a 3 and a 2, and 6 > 5. Similarly 6 can 3x3, 7 can be 3x4, etc. Basically, larger numbers are like using addition instead of multiplication, and multiplication will always grow faster.

[@MrConverse]

8:45, *product rule (not chain rule)

[@VinaySingh-jm8iw]

If you expand this problem include real numbers, the strategy changes from "use 3's and check if the excess is less than 2" to "use e's and check if the excess is less than e/(e-1)"

[@yurenchu]

Less than e/(e-1) ? How so? If real numbers are allowed to partition 1976 , the maximum product is actually a product of rational numbers, namely (1976/727)^727 ≈ (5.03024188) × 10^315

[@rohangeorge712]

i solved this problem intuitively as well. its a beatiful logic problem. i love these types of videos.

[@aroundandround]

Nice, without a pen and paper, I intuited that if we had to break the number into exactly two parts, breaking even numbers evenly and odd numbers off by one is the best strategy, so I figured we might be able to recursively use that strategy to keep breaking them further into roughly equal parts, but couldn’t think of why that would necessarily maximize the product; learned from the video it wouldn’t (because we have to minimize the number of 2’s).

[@voronische]

funny to see so different approach (I haven't finished watching yet). Mine was -- we need to maximise a function (1976/x)^x. Derivative, =0, x=1976/e. So, the best option is not 3, it is e, but we need a whole number, so it is 3 then. But theoretically, the best option would be e's.

[@voronische]

looks like you are getting right there :)

[@Tsunami1972]

Without knowing the rule regarding e, I just looked at products of numbers totalling 6, 12, and higher which were sums of 2s, 3s, 4s, and 5s. The product to sum ratio peaked at 3x3. 2x2x2 is 8, 3x3 is 9. Going higher 4x4x4 is 64 while 3x3x3x3 is 81. And it gets worse the higher you go. So, it's to the highest power until the sum is less than 1976, then x2 for the remainder, so 2x3^658. If it was 1975, it would change slightly to 2x2x3^657 because 2x2 > 3x1.

[@ghassanebentahar5695]

Amazing

[@seedmole]

Using Puredata, a visual programming environment for math/audio/data processing/etc... I ended up with the following expression: max($f1 * ($f2 - $f1), $f3) where $f1 represents one of the partitioned values, $f2 is the given unpartitioned sum, and $f3 is the output of the expression (with a starting value of zero). By setting the value of n at $f2, and sweeping the values across $f1, the answer is produced at the output (and will remain as the value of $f3 until the function is either reset, or both a new $f2 is set and values are swept across $f1 again. This can then be expanded to include more and more partitioning terms between $f1 and $f2, which in puredata could be done dynamically via abstractions, cloning, or creating new objects parametrically.

[@nathankarslake9351]

When I saw the equations with S, x, y and ln x as variables, I thought somehow the “surprising answer” was going to include Sexy. NGL, slightly disappointed it didn’t

[@hhergert78]

But it kinda is - 3 is "leet-speak" for "e"... or if you prefer to be more mathematical, the reason that 3 is the answer is that the optimal solution (if reals were allowed) would be the "e" that you're missing😉

[@TheEternalVortex42]

In general e is not going to actually give you the maximum in the continuous case. This is because you have to take a whole number of copies (obviously it still works if you can take a continuous "number" of copies but that doesn't really make sense then). So it's always better to take N/ceiling(N/e) as your number instead. For example if N = 32 then taking e you get e^11*2.09 = 125669 vs. (8/3)^12 = 129307. Moreover, we can make e an arbitrarily bad choice by picking N such that the fractional part of N/e is as small as possible. For example if you take N = 11 then using e you get e^4*0.047 = 0.71 which is quite bad! So the proof at the end doesn't actually work as written.

[@cherkovision]

I thought of a similar problem: What two numbers that multiply to 24 are greatest when one is raised to the power of the other? In other words, which of the following is greatest: A) 1^24 B) 2^12 C) 3^8 D) 4^6 E) 6^4 F) 8^3 G) 12^2 H) 24^1 Intuitively, you'd think that since the first and last options are obviously the smallest, the answer must be either D or E, or maybe B. And yet, the answer is C. There's something powerful about 3.

[@NeckMover61]

Very nice problem. Would be interesting to analyze it for the general case, see if 3 will also appear in the answer there.

[@cherkovision]

@@NeckMover61 As far as I can tell, for every number X that has 3 as a factor, the two factors that multiply together to give you X that result in the greatest number when one is raised to the power of the other is always 3^(X/3).

[@yurenchu]

​@@cherkovision ... unless X is non-positive, of course.

[@cherkovision]

I woke up last night thinking about this: As long as N is greater than 0, 3^N is greater than (or equal to) N^3. 3 is the only integer that this is true for. I can't be the first person to have realized that. Is that a thing? Does it have a name? Edit: I asked ChatGPT, and it told me that this has something to do with the number e (2.718). Since 3 is the closest integer to e, then if you're only comparing integers, 3 is the winner. However, e is the actual magic number (e^N is always greater than N^e).

[@micknamens8659]

Because 3 is a divisor of 24, your task is an application of the Olympiad problem solution, case 1, k=8. 3+3+3+3+3+3+3+3=n=24, 3×3×3×3×3×3×3×3=P.

[@joelcon4982]

Lol I thought it had to be just 2 integers for some reason and I was like damn that's easy it's just 988*988. I am delighted to be wrong.

[@virdrae]

Spotted it fairly easily. Should I take that as a good sign for the upcoming qualifiaction round?

[@KudoShinichii1412]

ngl I got lost and did not understand what just happend

[@Abdul_the_next_door]

It is mind-blowing how calculus popped up even in number theory.

[@erikkonstas]

I mean, not exactly, because the proof doesn't involve calculus at all, it's the "intuition" that uses it.

[@ryanburkett949]

@@erikkonstas True. But you can totally prove it with Calculus. After seeing this video I am going to do this problems as one of my "problem of the week" bonuses in AP calculus after we derive "e" with limits. I will hint them in the right direction but I hope they make the connection.

[@TheEternalVortex42]

There's a whole field called "analytic number theory"

[@ouwebrood497]

3 is the only integer between e en pi. It's one of my favorite numbers.

[@PrometheusMMIV]

My guess was pretty close. I was thinking you would want as many 2s as possible, which would give you the highest exponent. And possibly one 3 if the number is odd since it's better than wasting the 1.

[@michaeledwards2251]

Maximise 3s, 6 = 2+2+2 product 8 or 3+3 product 9.

[@PrometheusMMIV]

@@michaeledwards2251 Yeah, I watched the video too

[@bjorneriksson2404]

Huh, I actually got it right! I felt that my reasoning was on the intuitive side and not stringent at all, but I arrived at exactly this solution. I started by concluding that n^2 > (n+1)(n-1), which means that you should want to divide the original number into equal parts as far as possible. Start by dividing the number in half, since the resulting square is greater than the original number. Then, for every "larger" number, continue to divide into (more or less) equal parts. Then I recalled that ln(x)/x is a decreasing function for every x>e (this is connected to the old question of "if a

[@derekhasabrain]

I did the same as you, but only the first half and I just left it at 2’s. I completely forgot that 3^2 is larger than 2^3, let alone remembering the characteristic you mentioned. That’s incredible!!

[@leif1075]

What made you think of ln x or natural.log at all. Doesn't seem like something that would be orga icallt connected or you'd think of it

[@bjorneriksson2404]

@@leif1075 Well, to give you a long answer... what I actually thought about was, if you continue to divide the original number into smaller and smaller pieces, how small should you really make the pieces to reach "max product"? And it was then I thought about this frequently asked question of whether a^b or b^a is bigger (when a=3, thus pointing me to the answer that it is better to divide the number in as many smaller pieces as possible, as long as the pieces are not smaller than 3, than to have fewer and larger pieces. In other words, for e.g. the number 12, it's better to divide it into four 3's (with product 3^4=81) than three 4's (with product 4^3=64). And if you divide into 2's, you get the same result (2^6=64) as when dividing into 4's, so it's still better with 3's. Anyway, the ln(x) thing comes from a common proof of a^b > b^a. You can log both sides (with any log base, but its common to use ln), which gets you b*ln(a) ? a*ln(b). Divide both sides by a*b and you get ln(a)/a ? ln(b)/b. Then set f(x)=ln(x)/x and find the maximum of f(x) by taking the derivative, in the same manner as in the video. Then you find, as stated in the video, the maximum when x=e, and that for every x>e, the function is decreasing. This proves that when ae, ln(a)/a > ln(b)/b, and equivalently, a^b > b^a. So, it was simply this question and its proof that I came to think about when I was thinking of the best size of each piece. Here's a🌹for you if you managed to not fall asleep trying to read all this... 🙂 PS. If you've not heard of "wolfram alpha", google it (it's a web site) and let it plot the curve of ln(x)/x if you want to see how it looks.

[@lisa22150]

My first thought was, surely it does with as many 2s as possible and maybe a 3. But I very quickly shot that down. So then I thought probably as many threes as possible but I didn't get any further before I unpaused. Just a fuzzy thought.

[@Bhawani143]

I am your fan sir you are amazing & I know you have work on mathematics ❤love from India, Rajasthan,bhilwara~ Jai shree RAM~

[@AstroTibs]

Wow, I actually arrived at this in a different way. I started with the knowledge that if you split a number into two numbers, their product is maximized if the split is equal. I then assumed, on a hunch (not knowing about the AM-GM inequality), that no matter how many splits you make, the product is maximized if the splits are all equal size. It then came down to figuring how how many splits were necessary for 1976. I figured it was probably equal to 1976/2, or 1976/e. Turns out it was the latter, indicating that 3 is the best split to make. I then dealt with the remainder the same way you did. In other words, I figured out the second half of your video first, and the answer spilled out from there.

[@laurendoe168]

The answer I came up with is VERY VERY wrong. It was, in essence (1976/4) ^4. ROFL For SOME reason, I incorrectly presumed two things: (1) The all the integers had to be different, and (2) the more integers there are, the lower the product would be (don't ask me why I presumed either of these things - I haven't a clue).

[@HopUpOutDaBed]

my dumbass thought we were only picking 2 integers, so I divided the number in half and felt like a genius for 2 seconds till I watched the rest of the video

[@aonodensetsu]

quick answer from experience for two numbers is P=floor(sqrt(n))*ceil(sqrt(n))

[@HemantPandey123]

AM GM is which I used. But the first method is more intuitive.

[@NemisCassander]

Interestingly, in constrained optimization (thinking of LP here), assuming that a solution to an integer programming question is the integer closest to the solution for the relaxed (continuous) version of the same problem is one of the first things they tell you _not_ to do.

[@olayinkaanifowose5099]

I randomly used 3000 as an example, got to 2^1500 < 3^1000, and just guessed from there lol.

[@Mike-H_UK]

Easier to say.... For n > 4, 2(n - 2) > n, so the maximum product cannot include any integer greater than 4. Also, 2^3 < 3^2, so it cannot include more than two 2s. Since 4 = 2*2, it cannot include both a 2 and a 4. It obviously does not include a 1, since n + 1 > n * 1. So the maximum product must be made up mainly of 3s, with either no, one or two 2s (or equivalently one 4). Hence, 1979 = 2 + 659*3, the maximum product is (3^659)*2.

[@-ZH]

Hey, I managed to get it right. Guessed it by seeing that 2x2x2=8 While 3x3=9 And 4x4=16 While 3x3x2=18 So I guessed that 3 was the most efficient number to use and got 3^658x2

[@-ZH]

Which makes sense too, since x2 requires 2 sum x3 (2x1.5) requires 3 sum And x4 is the same efficiency as x2

[@TheEulerID]

My intuition is that the answer is 2^988 (on the basis that 988 x 2 = 1976). That's (approximately) 2.6 x 10^297. Quite what the proof of this is (assuming I'm correct), I will have to think about, but the intuition is based on the observation that in such questions the value of increasing the size of the exponent generally overwhelms other considerations.

[@egoudoun5034]

I’d like to see the answer if we add the condition that the numbers are all different.

[@markvanderwerf8592]

My guess: It will be in the form of 2+3+4+5+6 etc. But to make a fit you'll have to kick out one number. For example 2+3+4... + n etc is 1/2n (n+1) - 1. So for 1976 closest is n=63 being 2015. Then you kick out 39 to make the sum 1976. And the product then is 63! / 39. Alternatively you can see it as taking 2+3+4...+62 (which is 1952) and then adding on the remainder of 24 to 39 which makes that term 63. Either way you're getting as many terms as possible and you're bringing them as close to 3 as possible.

[@sytrostormlord3275]

@@markvanderwerf8592 isn't the sum of 2+3+4+5+...+n = (n+2) (n-1)/2 ? where n+2 is the sum of lowest and highest number is series, and (n-1)/2 is the number of such sums? that way you get (n+2)(n-1) ? 3982 sqrt(3982) ~63, so that's a good starting point of finding n: for n=63, we get 65*61 =3965, and a leftover of 3982 - 3965 = 17 so, the sum, would be: 2+...62 + (63+17) which leaves the product of 2*...*62 * 80...= 62! * 80

[@Br3ttM]

There probably wouldn't be a formula for that, and even knowing to use smaller numbers, multiplying every number starting with 2 and counting up until it reaches your goal would probably leave a remainder, and figuring out the best way to use up that remainder would probably require a lot of testing.

[@yurenchu]

​@@sytrostormlord3275 You're using the number 3982 , which suggests you're solving the problem for N = 1991 instead of N = 1976 . Secondly, you're mixing up a few things in your calculations: for n=63, you wanted to calculate (n+2)(n-1)/2 = 65 * 31 (and compare it to N = 1991 or N = 1976), but instead you calculated 65 * 61 and continued with comparing the result to 3982 for some reason? Thirdly, instead of the product 2*3*4*...*62*(63+17) , it's better to have the product 2*3*4*...*45*46 * 48*49*...*62*63*(47+17) because it is larger. Both products consist of 62 factors whose sum is 2032 , but the difference is that the first product contains the factors 47 * 80 while the second product contains the factors 63 * 64 instead; and 47 * 80 = = (63.5 - 16.5)(63.5 + 16.5) = 63.5² - 16.5² < 63.5² - 0.5² = (63.5 - 0.5)(63.5 + 0.5) = 63 * 64 which shows the second product is larger.

[@disgruntledtoons]

Now add the requirement that the numbers be distinct integers.

[@GeoffryGifari]

Can the solutions to all past IMO and putnam problems be found somewhere?

[@MyOneFiftiethOfADollar]

I don't think so. I think they all vaporize into thin air after exam completion.

[@GeoffryGifari]

@@MyOneFiftiethOfADollar makes sense!

[@ansambel3170]

here's a more intuitive solution imo: consider 6 which can be either 2^3 or 3^2. this means that all 2 should be swapped to 3's in the solution. now lets see if higher numbers are better or worse than 3's: 4^3 < 3^4. we can generalise the comparison with x^3 / 3^x this can be shown to be decreasing after 3 (i'm too lazy to do it in the yt comment, derivative is easiest, but probably can be done w/o it) since it's decreasing and 1 at 3, all other integers are worse than 3, 2 and 4 are second best, therefore, solution is maximum 3s and some 2's to fill up the number.

[@iodboi]

hi Presh.. I have a question at @1:36; If 2 is partitioned (as integers), then we get maximum two partitions, Partition_1 is 1,1 and Partition_2 is 0,2 Their product is 1 and 0, consequently, the maximum value of Product is 1 (for n = 2) Similarly for every subsequent n, ONE value of the Product is (n-1)

[@iodboi]

Essentially, what I am saying is that 2 is not the maximum product. As partitioning 2 will yield 2 and 0, whose product is 0. Thus, the maximum product is 1

[@thomasmaughan4798]

Probably 2*2*2*... a list of 988 2's or 2^988 (two to the 988th power). The sum of the 2's is 1976 and the value of the products is 2.6x10^297

[@justpaulo]

I don't think you can say that 3=3 (n=n in general). You either say 3=3+0, but in that case the product is 0, the smallest value possible, so it doesn't matter, OR you say that 3+0 is not possible because 0 is not an option and then you don't have a valid sum in that case b/c there is no sum in 3=3.

[@pedrogarcia8706]

if that's the case then all the products are 0 because you can add 0 to all of them

[@justpaulo]

@@pedrogarcia8706 If 0 was an option (it's not since we're talking about positive integers) you still can choose not to add 0, the same way you avoid the +1. But I claim that in the case 3=3 (or in general n=n) you have no way to avoid it. You need to have +0 otherwise I think there is no valid sum on the RHS of the equation. But since since 0 is not an option, 3=3 is invalid imho. It doesn't changes the end result, but I think it's confusing and not nice to have n=n as a solution.

[@RGP_Maths]

​@@justpauloAre you saying that you have to have two or more integers for the sum to be defined? Because that is not the usual understanding of the word. The sum of all the numbers in the set {3} is 3.

[@justpaulo]

@@RGP_Maths I'm no mathematician. My definition of addition is more colloquial...

[@olerask2457]

Again the Euler number is the NATURAL choice 😂. Wonderful problem 😊.

[@MarcusCactus]

Three questions in one: 1) Max product(x1,...,xk) s.t. sum=N. x real, k fixed. -> Simple Lagrangian, all x equal. 2) Find best k. x real. -> Max x^[1/x] -> x=e=2.71828 3) Find a solution in integers. -> Lots of 3, one or two 2.

[@eschiedler]

In high school, I stumbled onto the proof that for any x^2, (x+1)*(x-1)=x^2-1 therefore for any even positive number, divide by two and you have x. To get the sum of two different integers, add and subtract one then multiply you will have the maximum possible number.

[@aayanshah6783]

What do we obtain from this?

[@poly_hexamethyl]

How about this for the original problem? Given a, find x and y such that a = x + y and b = xy is maximized. b = x(a-x), so b'(x)=a-2x which is 0 when a = 2x. So x = y = a/2 and xy = (a/2)^2. For a = 1976 you get x=y=988 and xy=976,144.

[@krzysztofmichasieroslawski2682]

If we take the numbers while two of them are negative. (-1)*([say]-6400)*([say]6400-[n+1])

[@poznyakpoznyak]

It is 2×3^(1974/3). In general, you don't want any number N>4 in your product, because you can replace it with 2 and (N-2) and get greater product. Obviously, you also don't need the number 1, because it contributes nothing to the product. So it's only 2's, 3's, and 4's. Now, any 4 can be changed into 2 and 2 without affecting the product, so it's just 2's and 3's. And any 2×2×2 is less efficient than 3×3, so you can't have more than two 2's. So the answer is always "take as many 3's as you can, and if needed, take a single 2, two 2's, or a 4".

[@MrConverse]

10:23, there appears to be something wrong with your text. Perhaps you meant to write “we pick the one closest to e…” but the word one is missing. Hope it helps! Thanks for the explanation.

[@megauser8512]

The words "one" and "to" are both missing there.

[@JavierSalcedoC]

There's way more demand than supply for small numbers

[@stevengordon3271]

Much more interesting question if the numbers which add up to N have to be unique (all different from each other). Not that I know the answer.

[@bkucenski]

3x + 2y = 1976 to maximize 3^x * 2^y

[@pulsar22]

I solved this simply by trial and error. 1st I tested whether I get a high P if I just divide n by 2 and square that. I got a low number. Then I tested if I instead raise 2 to n/2 which I got a good high number. Then tested for 3^(n/3)*(n mod 3), 4^(n/4) * (n mod 4), upto 5^(n/5) * (n mod 5) ... since i noticed that it keeps getting smaller after 3, I stuck with 3.

[@lucykitsune4619]

e is very aptly named because it truly is everywhere

[@wilhelmmeyer89]

nice.

[@jeffdege4786]

1979? Pre-internet? You may not have been using the Internet in 1979, but some of us were.

[@bg6b7bft]

Napkin math, I started with small numbers and worked my way up. Looks like you get the most out of "3". 1's are worthless. 3's are better than 2, because 2*2*2 < 3*3 4 is exactly 2*2 so meh. 5 as a multiple is worse than 2*3 6 is worse than 3*3 7 is worse than 3*4 8 is worse than 2*3*3 so I figure that 3*....*3 *2 or *4 based on the modulus is the best.

[@Aim54Delta]

Before watching the video... let me see if I understand the problem correctly. We are looking for the largest product number from any series of integers whose sum is 1976. That is to say - 1976/2 = 988, so 2 multiplied by 2 988 times is a valid answer for 2^988. Which is my intuitive answer for that question. However, integers do not exclude negative numbers, so we can actually develop an infinite series of negative and positive integers of any arbitrary value which sum to 1976, but when factored to a product exist in a domain of math theory I've not really dabbled much in. I suspect, though, that my interpretation of the problem is not what is intended - or that there may be a way of extracting higher values out of factorizations I am not applying. I guess this depends on what we are considering large as a number, too. I am thinking in terms of absolute integer value - but any way of achieving a non repeating, non terminating decimal would be arbitrarily large in terms of data needed to represent it. I have no idea how to do that with a product of integers, but I imagine some realm of math can deliver.

[@thomasmaughan4798]

The problem specified positive integers so as to avoid this possibility of an infinite series that sums to 1976.

[@perakojot6524]

I don't know what is so surprising about the answer. It is so well known that x^y is largest for x*y=const when x = e.

[@douwehuysmans5959]

2^(1976/2)

[@Spongman]

isn't that the product rule (not the chain rule) ?

[@TheEternalVortex42]

Technically the product rule is just a special case of the (multivariable) chain rule

[@TheKsh]

When i can't solve the questions like this, i just write a code to do it XD

[@user-jj8kg5ef2t]

Is there a general solution for a number S, that has 3 partitions, such that x+y+z = S and yield a maximum product P = xyz?

[@mastick5106]

The general solution would be to choose x, y, and z to all be equal to S/3 or the closest integers. For example, if S = 20, S/3 = 6.666... making the closest integers 6 and 7; 6+7+7=20 and P = 6*6*7 = 252. In general, we would have P = x*x*x OR x*x*(x+1) OR x*x*(x-1) if we perturb the system away from that state, we lower the product. If we change x*x to (x+a)*(x-a), we get x*x - a*a which is less than x*x. Changing x*(x+1) to (x+a)(x+1-a) gives us x*x + x + a - a*a which is either equal to x*(x+1) when a=1, or smaller for a>1. You get similar results for x*(x-1) or for changing which term has 'a' added and which subtracted.

[@yurenchu]

x+y+z = S are not "3 partitions" ; that's just _one_ partition into 3 parts. Three partitions would be like: 1+1+4 , 1+2+3 and 2+2+2 are three different partitions of 6 (into three parts).

[@justsomeboyprobablydressed9579]

Back when the IMO had questions I could solve.

[@micknamens8659]

2:32 Considering a partitioning with factors 1 is a waste of the viewers time when you're aiming for the maximum product.

[@fuxtnegrx]

Why do you substitute + and *?

[@piol29]

That proof with e rounded to 3... I remember from my studies, once professor told us that based on this theory, "trinary" computers should have been much more effective than binary, only technical restrictions caused that binary won... But it was 25 years ago, don't remember all details :-)

[@micknamens8659]

Each bit in a number doubles the maximum possible value P. But you need only 1, not 2 places to store the information for a bit. Hence the formula of the puzzle doesn't describe how to calculate the storage size. Btw. there are technologies for solid state memory (SSD) that use 3 or 4 different states instead of just 2. They are cheaper but also a bit slower than pure binary memory.

[@chezeus1672]

iirc, my prof told us ternary lost basically because of mccarthyism. the most famous ternary computer was built in 1958 for moscow state university, and it performed very well for its time. building a ternary computer in the west would have been political suicide at the time, and once it would have been possible again, higher programming languages were already common for binary computers. however, it's quite possible ternary is about to make a comeback, since quantum computers are going to require new languages anyway and the benefits of ternary don't just go away.

[@micknamens8659]

@@chezeus1672 Please describe in detail the benefits of using base 3 over base 2 for storing and processing information in computers. Otherwise it sounds like a religious belief.

[@chezeus1672]

@@micknamens8659 i'm not even arguing for or against ternary computers, and not being convinced either way couldn't be further from a religious belief. i would say i couldn't care less, but that's not true since ternary logic is a giant can of worms and i'm lazy. all i said is that binary won initially for political reasons, and that ternary is being explored again now due to the development of quantum computers from scratch. the potential benefit is obvious and couldn't be any simpler: 2^x < 3^x for x ∈ N, where x is the number of bits or trits. which means more throughput at lower bandwidths or frequencies, i.e. less heat. current quantum computers use superconductors to avoid all heat anyway (heat is a type of quantum information, it would change the data), so the benefit may be in the better usage of the available bandwidth, instead. it could reduce operating costs, it could make programming even more complicated for no benefit, or it could make the quantum computers exponentially faster. i don't know and while qubits seem to have the upper hand right now, it's too early to rule out the other option. unlike with higher bases, differentiating the 3 states -1(current in one direction), 0(no current) and 1(current in the other direction) is easy and robust, and here's the kicker: even binary, classical processors are inherently capable of doing that, they just lack the logic to use one of the possible states. until this point, it's really straightforward. the difficulties begin once you realize you'll have to replace boolean logic with something that can deal with a third option, i.e. new gates based on a less intuitive and more complicated system of logic. it would be an absolute mess.

[@omaraldebs8206]

❤❤

[@secondengineer9814]

When we say e is "closest to 3" are we talking arithmetically closest or geometrically closest? e-2 is ~.718... and 3-e is .281..., but also 2/e is ~73% and e/3 is ~90%. So what if we said "you can only choose between 2 and 3.6"? 3.6 is ~.881 away from e and 2 is .718 away. However, e over 3.6 is 75.5%! So the ratio of 3.6 to e is closer to 1 than the ratio of 2 to e! Say our sum was 18. We can either use 9 2s or 5 3.6s: 3.6^5 = 604.66 2^9 = 512 So it seems like the important part is that the *ratio* between e and 3 is closer to 1 than the ratio between 2 and e!!!

[@aaryan8104]

please correct me if i got the quesion wrong(im pretty sure i did) we can write 1976=988+988=n(2n=1976,using 1976 divided by 2 as there will be two sums and thus two multiples) every product will be in the form (n-x)*(n+x) as n-x=n+x=2n=1976 (n-x)*(n+x)=n^2-x^2 where greatest number will be infact when x=0 therefore answer is n=988 and number is 988^2 ?????

[@aaryan8104]

OH WAIT APOLOGIES i didnt understand the fact that there can be as many multiples

[@aaryan8104]

but if there were only two multiples id reckon this is how its supposed to be solved?

[@Adr-zo8es]

I found that if you take 24 (and some other numbers) and use 2.6666 (like you use 3 with whole numbers) you get a bigger product than when you take e. Can someone explain this?

[@Adam-gd6pp]

Think of it geometrically: (1) Given a fixed perimeter, the area of a rectangle is maximized when that rectangle is a square. (2) This generalizes to higher dimensions (replace perimeter, rectangle, and square with sum of edge lengths, rectangular hyperprism, and hypercube respectively). (3) Since 2.6666 goes into 24 exactly nine times, using 2.6666 corresponds to constructing a nine-dimensional hypercube. Using e corresponds to constructing a rectangular hyperprism that is slightly longer in eight dimensions and shorter in one. By (2), the hypercube has greater volume. QED

[@Adr-zo8es]

​@@Adam-gd6ppSo if I am correct I previously took the wrong way. The optimal solution is: "P = e^(n/e) ; P is the highest value possible for the number n ; n is the number of your choice".

[@Adam-gd6pp]

@@Adr-zo8es Yes, if P can be partitioned into a fractional number of fractional values.

[@TheEternalVortex42]

In general e is not going to actually give you the maximum. This is because you have to take a whole number of copies. So it's always better to take N/ceiling(N/e) as your number instead.

[@antaresmaelstrom5365]

Immediate thought: 2^988 on account of how exponential growth works. EDIT: Brainfart. For some reason the above quick thought was of the opinion that 2^3 was bigger than 3^2.

[@jeffdeupree7232]

I thought I understood the question going into this, but half way through I don’t have any idea what the question is.

[@TerenceRegan]

2^988. It doesn't ask for unique numbers.

[@stQZuO]

The argument of the final part is flawed: f(x) is maximum at x=e, doesn't necessarily mean in the case of integers, the maximum will be at 3. That is true if the curve of f(x) is symmetrical in the region close to x=e, but we don't know that. Actually, it could well be that although the f(x) is maximum at x=e, f(2) is still larger than f(3).

[@yurenchu]

He may have neglected to address it. But it's relatively easy to see that f(2) = f(4) , and since f(x) is continuous and has been shown to be strictly decreasing for x > e , we have f(e) > f(3) > f(4) = f(2) , hence f(3) > f(2) .

[@fabioberetz]

What if XiXj? (no repeated numbers)

[@jrok9996]

I'm guessing powerful functions like tetration, pentation, and hexation were off the table as representatives even tho they are short hand for multiplication... 3 hexated 1973 is a... large number...

[@michaeledwards2251]

I noted from other comments, the question was intended for 3rd graders, 8 to 9 years old.

[@djdoc06]

Interesting, came up with 3s by just extrapolating from small numbers starting from 8, 9, 10, 11…

[@tayserbinjafor1569]

Recently, I found a grammatical mistake. It's not "such that", rather it's "so that".

[@damiangruszka40]

1. you could get the derivative e.asier using x^{S/x}= e^{S1/xlnx). Gives the same result in one go. 2. To claim thet we take the closest intiger and that would solve the problem, is IMHO not correct. You should rather show that 3^(S/3} > 2^(S/2) for S>= 1. Which it is. since 3^{1/3} > 2^{1/2}.

[@moom2k829]

i found the question misleading. thought it meant find a and b that maximizes a*b, with the condition that a+b = 1979 edit : my bad, thought it said only two integers

[@RainbowDashShadesOfApproval]

2^988

[@Math_Rap_and_GOP_Politics]

So the true maximum value is e^(1976/e)≈ 5×10^315

[@huzefa6421]

Sir i would want your attention on this case pls. 🙏🏼 Most people recently, in my calss as well ( grade 10th) are not knowing how to solve equations like 5 - 3 + 8 Using Bodmas they are doing 3 + 8 first and then 5 - 11 = -6 Actual answer is 2 + 8 = 10 but idk for what reason are they simply messing up simple bodmas Do a community post 1st if possible on such a poblem ( where 2 signs are of same presedence. *-* , + Or *×* , ÷ ) and see which of the 2 possible answers are chosen more. like 5 - 3 + 8 i) 10 or ii) -6

[@cdmcfall]

They do not understand that addition and subtraction are solved simultaneously. They see "addition" before "subtraction" within the mnemonic and incorrectly believe that addition comes first. If you want to explain it to them, remind them that subtraction is simply a special form of addition that involves the additive inverse of at least one of the addends: a - b + c = a + (-b) + c. In other words, addition and subtraction are different forms of the same operation. Here is a more comprehensive order of operations, excluding unary operations like trig functions, principal square roots, and logarithms: 1. Brackets/parentheses/braces/other strong grouping methods like the vinculum (technically not an operation) 2. Tetration and other binary hyperoperations 3. Orders/exponents/indices 4. Multiplication by juxtaposition (weak grouping; not a universally accepted convention, which often leads to online debates) 5. Multiplication (including negation)/division (multiplication of the reciprocal or multiplicative inverse) 6. Addition/subtraction (addition of the additive inverse)

[@HopUpOutDaBed]

@@cdmcfall bro if they can't do 5-3+8 they aint gonna understand all that

[@123xqp]

When the operators are of the same precedence you apply them left to right. Think how you'd do it with money. Assume you have 100 coins and had to give person A 25 coins and person B 10 coins. You'd subtract 25 from 100 and then you'd subtract another 10, leaving you with 65. You wouldn't subtract 10 from 25 and then subtract 15 from 100.

[@cdmcfall]

@@HopUpOutDaBed I tend to get carried away. TLDR: Addition and subtraction are the same thing. 👍

[@huzefa6421]

​​@@cdmcfallI tried that but no one understands why exactly 😕 Especially the part where we have to include (-b)

[@trackinggod8087]

There seems to be at least one unmentioned rule. You apparently cannot use zero.

[@thomasmaughan4798]

You can use zeros, but all it takes is one of them to produce a zero product.

[@trackinggod8087]

@@thomasmaughan4798 My point was that you could have a 1 with 10000 zeros and easily have a larger result than any mentioned.

[@andrewkost2625]

988. 1976/2 = 988. This will yield the largest product. Idk how to prove it but I know for certain that if given 2 numbers x and y: x^2=y ALWAYS GREATER THAN (x-n)(x+n)=y-(n/2)^2 Would love someone to explain this to me

[@andrewkost2625]

this answer was made assuming only 2 factors, nevermind lol! my previous comment still interests me though