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Daily Putnam Problem

Anonymous Fri 04 Oct 20:49:36 2019 No.11032191 View Reply Original Report

Quoted By: >>11032832 >>11033043 >>11033140

What happened to the Putnam poster? I liked making futile attempts at solving them. Anyway, here's today's:

http://www.math.harvard.edu/putnam/index.html

Anonymous

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Anonymous Fri 04 Oct 2019 21:07:20 No.11032241 Report

Quoted By: >>11032250

Immediate thoughts:

Firstly, I think it must be a 2 & 3 split
(since the smallest 4 & 1 split, namely n (n + 1) (n + 2) (n + 3) - (n - 5) is always positive)

Can we say more about which two numbers must go together, and on what side?
Do we compare the products best using a difference to 0, or ratio to 1, or what?
Do we parametrize the difference somehow?

Anonymous

Anonymous Fri 04 Oct 2019 21:07:36 No.11032242 Report

Quoted By:

here's an idea for a brute force solution:
case 1: both subsets have 3 elements. Then for every choice you get an equation, for example
n(n+1)(n+5) = (n+2)(n+3)(n+4), which is easy to solve since n^3 vanishes and we're left with quadratic equation. That's 10 quadratic equations to solve.
case 2: subsets have unequal numbers of elements. Then
(n+5)(n+6) >= product with smaller no. of elements = product with bigger no. of elements >= n(n+1)(n+2)(n+3), so not much to do here.

Anonymous

Anonymous Fri 04 Oct 2019 21:11:27 No.11032250 Report

Quoted By:

>>11032241
>Firstly, I think it must be a 2 & 3 split
>when there are 6 elements in the set
Anon, I....

Anonymous

Anonymous Sat 05 Oct 2019 00:39:07 No.11032832 Report

Quoted By: >>11032839 >>11033112

>>11032191
There is no such $n$ .

There are three numbers in that set that are not divisible by $2$ .
No more than one of those three numbers could be divisble by $3$ and no more than one of those three numbers could be divisible by $5$ .
So there is a number in that set that is neither divisible by $2, 3$ nor $5$ .
Since this number doesn't have $2, 3$ or $5$ as prime factors, it must have a bigger prime $p \geq 7$ as a prime factor.
As a consequence, the product of the subset that number is put into also has $p$ as a prime factor.
But the product of the other subset doesn't have $p$ as a prime factor, since none of the other numbers have $p$ as a prime factor.
So the products can't be equal equal.

Is this really a putnam problem?
If so, it's the first one I've ever solved :)

Anonymous

Anonymous Sat 05 Oct 2019 00:45:36 No.11032839 Report

Quoted By: >>11032843

>>11032832
bigbrain

Anonymous

Anonymous Sat 05 Oct 2019 00:47:38 No.11032843 Report

Quoted By:

>>11032839
thanks, I spent way too much time on this though

Anonymous

Anonymous Sat 05 Oct 2019 02:53:56 No.11033043 Report

Quoted By:

>>11032191
For any n, there are 3 multiples of 2, 2 multiples of 3, and 1 or 2 multiples of 5 in the set.
For the products to be equal, each part must have a multiple of 5.
So there must be two multiples of 5 (n and n+5).
Represent these by o----o
One of these is even.
WLOG (you'll understand) let n be even
Marking the multiples of 2 and 5 gives o-o-oo.
The unmarked elements cannot both be multiples of 3 since they differ by 2.
So there exists an element of the set divisible by a prime, p, bigger than 5 (p>=7).
Since there are only 6 elements in the set, there is only one multiple of p in the set.
Therefore the products cannot balance and no such n exists.

Anonymous

Anonymous Sat 05 Oct 2019 03:31:06 No.11033112 Report

Quoted By: >>11033119

>>11032832
>since none of the other numbers have p as a prime factor
can you explain this please?

Anonymous

Anonymous Sat 05 Oct 2019 03:34:09 No.11033119 Report

Quoted By:

>>11033112
nvm i see it now

Anonymous

Anonymous Sat 05 Oct 2019 03:48:10 No.11033140 Report

Quoted By:

>>11032191
Based.

n is 1 mod 7
n is 0 mod 5
each of n,...,n+5 must be a product of only 2,3,5

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