Find a+b+c+d

This problem is from the Harvard-MIT Mathematics Tournament. It was a good way to ease myself back to making videos after a break.

[wafflaaar]

I did another method:

I decomposed the numbers in terms of prime factors:

165 = 5x3x11
120 = 5x3x2x2x2
64 = 2x2x2x2x2x2
88 = 2x2x2x11

note that 11 (and no other numbers) is common in 165 and 88, so let's take a = 11

remaining from 165, we find 5x3 (15) which can also be found in 120. so let's take b = 15

after that, we have 2x2x2 (8) cropping up 4 times, which implies that they will occupy c and d

adding them together, 11 + 15 + 8 + 8, we get 42

[iMíccoli]

Interesting fact i just realized is that 437 can be expressed as the difference of 2 perfect squares:

437=441-4
437=21²-2²
437=(21-2)(21+2)
437=19×23 :D

[sr]

I live in the UK and run puzzle evenings in a bar. Basically it’s a quiz where you need very little general knowledge. Most of the questions are mathematical but involve a lot of logic, usually integers or what I call integer fractions. I do have to throw in a few word based problems, just to keep certain customers happy. Your videos give me great inspiration. This is a great but I will need to adapt it. Thanks

[Rai_Te]

I actually found the solution to split the four numbers (64, 88, 120, 165) into their primfactors easier here, bacause the only four possible factors (as combination of the forementioned primefactors) became immediately obvious.

[franolich]

Love your elegant solution. As noted in other comments, one can also solve for a, b, c, d first:

64 = 2^6
88 = 2^3 x 11
120 = 2^3 x 15
165 = 11 x 15

W.l.o.g let ab=64 ({ab, bc, cd, da} and (a+b+c+d) unchanged by a cyclic permutation of (a, b, c, d))
=> a=2^(6-k) and b=2^k some k∈{1, ..., 5}
Note cannot have k=0 or 6 since that would mean one of the other pairwise products would have share a factor of 2^6 which they do not.

Since b is a power of 2, bc must involve a power of 2. So there are 2 cases:

1) bc = 88 = 2^3 x 11 = 2^k x c
=> c = 2^(3-k) x 11
=> cd must have a factor of 11
=> cd = 165 = 11 x 15 = 2^(3-k) x 11 x d
=> k=3, a=b=8, c=11 and d=15
=> da = 15 x 8 = 120
=> (a, b, c, d) = (8, 8, 11, 15)
=> a+b+c+d = 42

2) bc = 120 = 2^3 x 15 = 2^k x c
=> c = 2^(3-k) x 15
=> cd must have a factor of 15
=> cd = 165 = 11 x 15 = 2^(3-k) x 15 x d
=> k=3, a=b=8. c=15 and d=11
=> da = 11 x 8 = 88
=> (a, b, c, d) = (8, 8, 15, 11)
=> a+b+c+d = 42

[nikhilprabhakar]

I thought the problem is nice but your solution is even nicer. Thanks


PS: I solved it by finding factors in each number and matching those which share some factors. 8 x 11, 11 x 15, 15 x 8 and 8 x 8
The answer is 42 but method is foolproof.

[aidenbooksmith]

I have been watching you for about half a year. And I am proud to say you taught me well. Before I began watching you, I was a high school math failure. Now, I finally managed to solve one of your questions on my own before watching your video!

[Grecks]

Like many others here, I solved this puzzle by looking at the prime factorization of the dual products. The principal idea was to understand that when for example ab = 64, then a and b must both be 8, and cd must equal 165=11*15.

I want to share my views on the inter-dependency of the number of solutions and the order of things given in the question. When being asked only for the sum a+b+c+d, then clearly the order of (a, b, c, d) does not matter, whereas the order of assignments of the 4 numbers {64, 88, 120, 165} to the products {ab, bc, cd, da} could theroretically result in more than one sum a+b+c+d for the answer, so we must carefully consider it. It turns out, though, that this is not the case here, and we're left with only one sum (42) for the answer. This is probably due to the cyclic symmetry in the selection of the 4 two-products.

On the other hand, if we ask for all the tuples (a, b, c, d) that fulfill the requirements of the puzzle, we arrive at 8 different solutions in total! Interestingly, this means that only one third of all 24 possible permutations of (8, 8, 11, 15) qualify as an answer.

[aavalos]

By inspection, you can note that 165 has to be the product of two odd numbers. From our multiplication table, even*even = even. Even*odd = even, odd*odd = odd. Lasty, 64 is a power of 2. Thus any two numbers that multiple to it must be a power of two. So we know two of the numbers are powers of two. This makes it easy to eyeball the solution based on the prime factorizations of the other numbers.

[pojuantsalo]

This was a cool number puzzle to solve! This is how I solved this. I started writing the given numbers 64, 88, 120 and 165 as product of their prime factors:
64 = 2*2*2*2*2*2
88 = 2*2*2*11
120 = 2*2*2*3*5
165 = 3*5*11

We have four positive integers a, b, c and d so that ab, ad, bc and cd gives us those four integers above. Each integer a, b, c and d appear twice in the pairwise products. We can instantly see that the prime 11 appears twice (as factor of 88 and 165). Since 3*5 and 2*2*2 don't have common factors, we know one of the numbers a, b, c and d must be 11. Let's set:

165 = 3*5*11 = ab => a = 3*5 = 15 and b = 11.

After this it is clear that:

88 = 2*2*2*11 = cb => c = 2*2*2 = 8.

Finally we solve for d:

120 = 2*2*2*3*5 = ad => d = 2*2*2 = 8.

We also see that these values check up for the final number 64:

64 = (2*2*2)*(2*2*2) = 8*8 = cd.

Solution: a+b+c+d = 15+11+8+8 = 42.

Clearly my way of solving this is different from the video, but my brain functions this way, so this is how I solved this.

[Sigma.Infinity]

This is a technique I've seen used a lot on this channel. Convert the problem into the form: (...)(...)=integer, then test each possible pair of multiplicands for validity.

[smalin]

88 and 165 are each multiples of 11, so d=11. 120 and 165 are each multiples of 15, so c=15. The greatest common factor of 88 and 64 is 8, so a=b=11.

[xenumi]

Simplify by picking it apart: 2 products have 11 (prime) as a factor, so 11 is one of the letters, while 8 and 15 are the two numbers that the first pairs with. 15 doesn't divide 64, so it must be one of the pair making 120 and it must be another of the letters. 120/15 is 8, so 8 and 8 are the remaining 2 letters. 11 + 15 + 8 + 8 = 42. Many of the comments are similar to this. Prime's solution seems like it would work better for less obvious variations of this problem.

[davidellis]

437 = 441-4, a difference of two squares. 21^2-2^2 = (21+2)(21-2) = 23*19.

[jbglaw]

I actually started this by cutting the four numbers into prime factors. From there, the answer is obvious as well.

[niraa]

A solution that wouldn't work with different numbers but still works here
64 is either 8² or 64x1 so we 100% get 2 of the 4 numbers, then by dividing them with the other numbers we get the remaining 2.

[RahulKumar-idcq]

Ur solution is really excellent.
But I solved in following way:
multiply all so we get.
(abcd)²=64*88*120*165;
abcd=15*11*8*8;
So following statement can be concluded:
Possible pair as (ab, cd) or (bc, ad)=(64, 165)or(88, 120);
we can also observe that
ab/bc=(cd/ad)-¹;
Case 1
Let ab=64 & cd=165;
bc=88 & ad=120;
64/88=(165/120)-¹=8/11=a/c ;
let a=8k, c=11k;
so b=64/8k=8/k
similarly d=15/k
as b=8/k and d=15/k
So maximum value of k will be HCF(8, 15) that is equal to 1.
So only value of k is 1
a=8, b=8, c=11, d=15
Case 2
Let ab=64 & cd=165;
bc=120 & ad=88;
Similarly a=8k, b=8/k, c=15k, d=11/k
again only possible value of k=1
a=8, b=8, c=15, d=11.
Case3
let bc=64 & ad=165;
cd=120 & ab=88;
Substitute in case 2
a=b
b=c
c=d
d=a
Case 2 become case 3.
Similarly all other cases can be concluded.

[HeartlessConservativ]

Prime factorization is another way to solve this.
64=2^6, so two of the unknown numbers are only powers of two and the sum of those powers is six.
165=3*5*11, so two of the numbers have no power of two.
88=2^3*11
Since we know that all four of the numbers are either powers of two or have no power of two, We know that the two numbers whose product is 88 are 2^3=8 and 11. This, then, allows us to determine the other two numbers from the remaining factors of 64 and 165: 8 and 15.
So the four unknown numbers are 8, 8, 11, 15, whose sum is 42.

[AndrejPanjkov]

List each of the four numbers as products of two factors..
Remove those products where the factor appears in no other list e.g. 165 = 33x5 cannot be one of the pairs, as 33 appears in no toher pair.
This leaves 64 = 8x8, 88 = 11x8, 120 = 15x8, 165 =15x11.
So a, b, c, d are some permutation of [8, 8, 11, 15].
The sum is 42.

[KRO_VLOGS]

Sir you should do a video on proving power rule by the definition. Its really complicated in our textbooks something to do with the binomal theorem