*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
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.
@@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.
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.
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
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.
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.
@@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
@@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
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.
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.
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.....
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)"
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
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.
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).
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
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 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.
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.
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.
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 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).
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).
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).
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
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)
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
@@iodboi That's the case when you partition the number 2 into _two_ terms. But we can also "partition" the number 2 into just _a single_ term. Just as a summation ∑ a[k] from k=m to k=n , can also contain only one summand (namely when m=n).
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
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.
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.
@@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
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.
@@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.
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".
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...
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 If a given even integer 2n must be written as the sum of two integers a and b such that their product a*b is maximized, then that maximum is realized when a = b , and hence a = b = n = (2n)/2 . (In the particular case that 2n = 1976, then a = b = 1976/2 = 988 .) Proof: Since 2n = (a+b) , we have n = (a+b)/2 is the mean value of a and b . Suppose a≤b , and let d = (b-n) = (n-a) = (b-a)/2 . Then a = (n-d) and b = (n+d) , and hence the product a*b becomes a*b = = (n-d)(n+d) = n^2 - d^2 which (for fixed/given value of n) is maximized when |d| is minimized. In particular, when d = 0 then a = (n-d) = (n-0) = n and b = (n+d) = (n+0) = n , hence a = b.
I hadn't watched this channel in years due to the explanations of the solutions often being fundamentally incorrect. I thought I'd have another quick look, as the problem in the thumbnail was easy enough to solve in a few seconds. I was somewhat surprised to see that the explanation is once again riddled with incorrect statements. I'm really sad to see that this channel has so many followers.
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
i feel i don't understand the problem correctly. if i simply divide 1976 by 2, i get 988. this means that i can construct a sum (2+2+2+2+2...) until its value reaches 1976, and i will have exactly 988 twos in that construction. if i multiply them all together, i get 2^988. so summing 988 quantities of 2 yields 1976, and their product yields 2^988, not 2^376
I divided into 2s at first too. But then i realized that I could swap every 2+2+2 by 3+3 and get a larger product, because 3*3 > 2*2*2 . So the largest possible product of positive integers whose sum is 1976 , is 2*(3^658) . (Note that 2 + 3*658 = 1976 .)
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 :-)
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.
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.
@@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.
@@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.
But, 658x3=1974 and 2x1974=3948 The question says nothing about raising one of the integers to a power of one of the other integers! - it asks for the maximum product from the integers that add up to the given number. My answer was 1111+865=1976 since 1111x865=961,015.
The correct answer is P = 2 × 3 × 3 × 3 × ... × 3 , with exactly 658 instances of the number 3 in the product; because 2+3+3+3+...+3 = 2 + 658×3 = 1976 . This product can also be written as 2 × 3^658 ; that's where "raising to the power" comes from. The value of this product is about 1.76528813 × 10^314 (it is an integer number that has 315 digits). By the way, if we're limited to only _two_ numbers whose sum is 1976 , then the maximum product is P = 988 × 988 = 976,144 .
I don't get it. What i did is the following: x+y=1976 y=1976-x I want to maximize x*y x*y=x*(1976-x)=-x2-1976x It is a cuadratic ecuation with a=-1. It has a maximum. I derivate, (-x2-1976x)'=-2x+1976 -->xmax=988 Then y=988 Then x*y=976.144
It didn't say _only two_ integers. For example, I could partition 1976 into three positive integers: 1976 = 1000 + 100 + 876 . The product of those three terms would be 1000*100*876 = 87,600,000 , which is greater than your 976,144 . Now, what is the optimum number of positive integers in which we can split up 1976 , such that their product is maximal? That's the question that is being asked.
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.
But what would happen if the natural logarithm base was different? let's say... 4.798 ? This does not mean that the answer can be "get as much 5s as possible", right? This video begs the question: "Why is natural logarithm's base is number 'e' ? " Can you make a video on that please?
Disregarding absolutely everything said in the video I approached the problem completely differently. We were asked to come up with largest product. To get large product we need large numbers. What are the largest numbers that make up 1976? They should be close to half. 1976 = 988 + 988 988 x 988 = 976144 If we try any other combination: 987 x 989 = 976143 986 x 990 = 976140 So the biggest products are numbers closest to the half of the original.
I determined within some 60 seconds 976135 being the largest number of a product the sum of which is 1976... ...actually you get 991 times 985 being the product and my first notion would have been 1000 times 976 but that gets out smaller the the first product... ..so just don't jump to conclusion.... Le p'tit Daniel, my mental calculation got me there... ...and I want desperately to be taken away from out of the dark to ponder also some algebraica again...
Hum, not sure I agree with the conjecture show at 4:30. At this stage in the presentation, you've demonstrated that product of 2 cubed is smaller 3 squared, thus using multiplication of 3 is more efficient and yield a higher product than product of 2. You've also demonstrated that products of 4 are just as bad as 2. But you've not yet demonstrated that products with a primary number larger than 3 are more efficient than 3. (5,7,11,13...) Maybe my math knowledge is flawed and for any mathematian that is an obvious answer. From where I am sitting. It is not. Maybe I should just watch the rest of the video?
How about if you use the n(n-1) / 2 method to determine the number of consecutive integers that will add up to 1976. And the factorial of the top number, which is the product of all the numbers together, would be a very large number... That doesn't prove it, but it's one way to get a very large number, even with no repeats, which wasn't specified. 1 +2+...+63=1953, CORRECTION: 1 +2+...+62=1953 1+2+...+6=21, 1953+21+2=1976 So total is at least 63! * 6! * 2 = 2.855 x 10^90. CORRECTION So total is at least 62! * 6! * 2 = 4.53*10^88
Your solution does have repeats: your product contains the factor 6 two times, the factor 5 two times, the factor 4 two times, the factor 3 two times, the factor 2 three times, and the useless factor 1 two times. (Note that 1+1+2+2 = 3+3, but 1*1*2*2 = 4 < 9 = 3*3 .) Moreover, your calculation contains an error: 1+2+3+...+63 = 2016 , _not_ 1953 . If you want a solution with _no_ repeats, the optimal partition would be 1976 = (2+3+4+...+37+38) + (40+41+...+62+63) as these terms give the maximum product (2 * 3 * 4 * ... * 37 * 38) * (40 * 41 * ... * 62 * 63) = 63! / 39 which is a 86-digit number. (However, the solution of the video, which utilises repeating 3s, is a 315-digit number, and hence much bigger.)
1000 x 100 x 100 x 776 would already result in a much bigger product than 988 x 988 . The challenge is to find the optimal number of terms, that result in the maximum possible product.
Since it asks for the product of each natural number chosen, 1^1976 = 1. It actually ends up being the smallest possible choice for the parameters of the question 🤔
@@Adr-zo8es no, i meant like 10^1976 but with 1's instead of 0's (i think it might be written as 10^1976 + 10^1975 + 10^1974... + 1). like the final positive integer is what i'm talking about. unless i'm misreading the problem the digits of the final number are not being multiplied they are just being added.
I see how I'm misreading the problem now. "Determine the largest number that is the product of positive integers whose sum is 1976" the "whose" confused me. I basically read it as "what is the largest positive integer whose digits' sum is 1976?" my bad there.
@@aboi5 , the number of 10s that would sum to 1976 are only 197 (and then add 6), not 1976. So you would get 10^197, which is far less than 3^659. In fact, 10^197 is .00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000003% of 3^659.
The maximum product is P = 2 * 3 * 3 * 3 * ... * 3 in which there are exactly 658 instances of the number 3 (because 2 + 3 + 3 + 3 + ... + 3 = 2 + 658*3 = 1976 ). That product can also be written as P = 2 * 3^658 . The value of that product is about (1.76528810...) * 10^(+314) , it is a 315-digit integer number.
off the top of my head I'd say that number is 988x988 (which my calculator says is 976'144) - and for the second one I'd go with 989x990 (979'110) now let's watch the video and see why I'm wrong 😂 edit: so apparently it wasn't necessary for me to see the whole video. my first mistake was to think I could only multiply 2 numbers.
Okay, before watching the answer, I'm going to guess that it's 2^(1976/2) ~= 2.615988x10^297. Let's see if I'm way off. Edit: intuition good, but I should have put a bit more thought into it.
"Well, you could take that number that's greater than 4, and you can split it into two numbers, so we have the number minus 2 and 2." What? I'm not following your grammar.
Well, you could take that integer number M which is greater than 4 (i.e. M>4), and you can split it into two numbers (M-2) and 2. Note that the product of (M-2) and 2 is greater than M , since M is greater than 4. Proof: (M-2)*2 = 2M - 4 = M + (M-4) Since M>4, we have (M-4)>0, hence the outcome M+(M-4) is greater than M.
This is a most interesting and wonderful chanel. But... may I ask something? I am always annoyed by the LOUD SHOUT, intended to be ....a salute...., at the beginning. There seems to be no reason for this shouting, too. A normally spoken "Hallow" or "Hey.." will be much better. Damn! This is probably a culture thing. And my simple demand will no be heard.
Before watching the video, here is my solution: 976144 Show my work: I'm guessing that I want the largest factors possible to get the largest product, which would be half of the starting number. Find two positive integers that sum to 1976. Since 1976 is even, I just divide it by 2, getting 988. Odd would be the floor and ceiling of the half. 988 + 988 = 1976 988 × 988 = 976144 The rules never said that the positive integers had to be distinct. Now to watch the video to see how close my solution is... ------- Not even close. 2 × 3^658 So, approximately: 1.7652881309 × 10^314 But at least I tried, which is probably more than most people watching the video did. 😂
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)
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,…
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.
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 😭
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 .)
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
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😉
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.
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.
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) .
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.
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.
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.
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.
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!
The result of the product cannot be e^(1976/e) , because 1976 must be the sum of an _integer_ amount of terms, even if the terms themselves don't have to be integers. (Note: 1976/e is not an integer, so you cannot have "1976/e copies of e" as factors of the product.) The highest product that we can have, if the number 1976 is allowed to be partitioned in non-integer real numbers, is (1976/727)^727 which is about a factor 30 greater than the 2*(3^658) answer in the video. However, no matter how we partition the number 1976, the product can never be greater than e^(1976/e) .
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.
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 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.
@@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.
@@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.
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?
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
@@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".
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.
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.
