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

- Thread starter eddybob123
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The usual terminology isperfect divisorif $\gcd (d,\frac{n}{d})=1$. Let $m$ be the sum of all perfect divisors of $2014!$. Find $m\,(\text{mod}\, 2014)$.

$$

\prod_{p\le2014}p^e+1

$$

where the product is over primes and the exponent e is the p-adic valuation of 2014: $e=\left\lfloor\frac{2014}{p}\right\rfloor+\left\lfloor\frac{2014}{p^2}\right\rfloor+\ldots$.

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:

Code:

```
val(n,p)=my(t);while(n\=p,t+=n);t
t=1;forprime(p=2,2014,t*=p^val(2014,p)+1);t
```

Not that I understood what you did... Anyways, it's your question!Hence the product is divisible, at least, by 2 * 19 * 53 = 2014 and so the answer is 0 mod 2014.

Hmm, maybe I didn't explain it well enough.Not that I understood what you did...

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?Anyways, it's your question!

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

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