eddybob123

Let a divisor $d$ of a positive integer $n$ be called a perfect divisor if $\gcd (d,\frac{n}{d})=1$. Let $m$ be the sum of all perfect divisors of $2014!$. Find $m\,(\text{mod}\, 2014)$.

v8archie

2014 = 2 x 19 x 53

Since all exponential of each prime factor is 1, every proper divisors of 2014 (i.e. not 2014 and 1) is a perfect divisor.

So m = 2 + 1007 + 19 + 106 + 53 + 38 = 1225 = 1225 mod 2014

CRGreathouse

The usual terminology is unitary divisor. The sum of the unitary divisors of 2014! is
$$
\prod_{p\le2014}p^e+1
$$
where the product is over primes and the exponent e is the p-adic valuation of 2014: $e=\left\lfloor\frac{2014}{p}\right\rfloor+\left\l floor\frac{2014}{p^2}\right\rfloor+\ldots$.

1861 is prime and divides 2014! exactly once (since 2014/2 < 1861 <= 2014) and so the product is divisible by 1862 = 2 * 7^2 * 19. Similarly, since 1907 is prime the product is divisible by 1908 = 2^2 * 3^2 * 53. Hence the product is divisible, at least, by 2 * 19 * 53 = 2014 and so the answer is 0 mod 2014.

For those interested in the nonmodular answer in its 5782-digit glory, here's a GP script:

v8archie

Somehow I read that "!" as English punctuation! I thought it was a bit easy (although I woulld in the cold light of day add 1 and 2014 to my answer).

eddybob123

Not that I understood what you did... Anyways, it's your question!

CRGreathouse

Hmm, maybe I didn't explain it well enough.

Since you want unitary divisors only, the basic units are not the primes but the prime powers maximally dividing the number. So let N = q1 q2 ... qk where each qi is a prime power and no primes are reused. Then for each qi it can be included, or not, in a unitary divisor. Thus the number of unitary divisors is 2^k.

To get their sum you could examine each of these 2^k numbers, but you can do better by noticing that the unitary divisors are either divisible by qk, or not, and you can transform one group into the other by multiplying/dividing by qk itself. So if you knew the sum of the divisors of N/qk you can multiply it by 1 + qk to get the sum of the divisors of N. Induction gives you what you;d expect: you can just multiply qi + 1 for all prime powers qi || N (that is, all p^e | N where p^(e+1) does not divide N).

Hmm. Draw a regular hexagon and connect all pairs of points. How many triangles are there in total?

v8archie

Does this include joining two triangles to make another? Or indeed, joining shapes of any type to make triangles?

CRGreathouse

Any triangle as long as all three sides are sides or diagonals of the regular hexagon.

v8archie

But triangles with sides that are segments of a diagonal are not included?

CRGreathouse

Those are included, too. (I was looking at everything as a line; if you look at segments instead you should use those. It all comes to the same thing.)