/math/ - Mathematics

*was wandering what

Damn proofreading...

The sum of n squares from 1 to n can be written as n(n+1)/2, so that must be a multiple of 200, or equivalently n(n+1) a multiple of 400. Let n=400 and we're done.

>>320

I'm terribly wrong.

We must find n*(n+1) such that their prime factorization contains 2^4 and 5^2, by the chinese remainder theorem i find n=175 to be the smallest such integer.

>>320

But n(n+1)(2n+1)/6 is the formula for the sum of the first n squares.

>n(n+1)(2n+1)/6 = 200k

>n(n+1)(2n+1) = 1200k

At this stage, we know that 1200 is an upper bound for the solution.

1200 = 2^4 * 3 * 5^2

>>322

You're right, i confused the formula for the sum of the first n naturals.

The question is about distributing the prime factorization of 200 over the three elements of our final product. I am not sure how to do that systematically.

Letting n=0(mod25) and n+1=2^4*3(mod48) i find a small solution of n=575.

If we impose a restriction to the term 2n+1, I suspect that will make n grow larger. So if we can limit our restrictions to the n and n+1 terms, we can make n smaller.

112 is the answer, but I can't find a way to do this without checking all values of n with a computer. Interesting problem though.

>>322

>>323

>>325

I don't have a direct solution without a search, but the search here is over factors, not values of n

Masonic dubs...

Must post...

n(n+1)(2n+1) = 1200 does not guarantee that n is an integer

the construction of a set of solutions based on certain members, does not guarantee the solution of that set has those members

it does imply that n(n+1)(2n+1) is 1200m, a multiple of 1200

1200 is 2^4 * 3 * 5^2

1200m = 2^x*3^y*5^z*m

m can be anything that has factors spread across the three expressions

observation:

6 and 2 are even

3 and 1 are odd

even * any = even

even + odd = odd

powers of 2 are restricted to either n or n+1, at least 4 times total

powers of 3 can appear in all of them, at least once

powers of 5 can appear in all of them, at least twice

2n+1 must be odd, so for 2n+1, the power of 2 must be 0

i = n, j = n+1, k = 2n+1

so this is how far I got

n = 2^x*3^y*5^z*u, n+1 = 2^p*3^q*5*r*v, 2n+1 = 3^s*5^t*w

2^x*3^y*5^z*u = 2^p*3^q*5^r*v - 1 = (3^s*5^t*w - 1)/2

1 = 2^p*3^q*5^r*v - 2^x*3^y*5^z*u

1 = 2^p*3^q*5^r*v - (3^s*5^t*w - 1)/2

2 = 2*2^p*3^q*5^r*v - (3^s*5^t*w - 1) = 2*2^p*3^q*5^r*v - 3^s*5^t*w + 1

1 = 2*2^p*3^q*5^r*v - 3^s*5^t*w

>powers of 2 are restricted to either n or n+1, at least 4 times total

>powers of 3 can appear in all of them, at least once

>powers of 5 can appear in all of them, at least twice

I failed that

if 1200 is the target form, then 2400, 3600, 4800 are viable tests

the highest power of 2 is 4

the highest power of 3 is 1

the highest power of 5 is 2

everything else is a multiplier

back later