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June 18th, 2014, 08:54 PM   #1
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Math Q&A Part 2

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 $m\,(\text{mod}\, 2014)$.
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June 18th, 2014, 10:09 PM   #2
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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
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June 18th, 2014, 10:24 PM   #3
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Originally Posted by eddybob123 View Post
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 $m\,(\text{mod}\, 2014)$.
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:
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
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June 19th, 2014, 05:26 AM   #4
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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).
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June 19th, 2014, 01:18 PM   #5
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Hence the product is divisible, at least, by 2 * 19 * 53 = 2014 and so the answer is 0 mod 2014.
Not that I understood what you did... Anyways, it's your question!
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June 20th, 2014, 06:29 AM   #6
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Not that I understood what you did...
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).

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Anyways, it's your question!
Hmm. Draw a regular hexagon and connect all pairs of points. How many triangles are there in total?
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June 20th, 2014, 06:40 AM   #7
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Does this include joining two triangles to make another? Or indeed, joining shapes of any type to make triangles?
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June 20th, 2014, 06:47 AM   #8
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Any triangle as long as all three sides are sides or diagonals of the regular hexagon.
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June 20th, 2014, 06:50 AM   #9
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But triangles with sides that are segments of a diagonal are not included?
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June 20th, 2014, 07:05 AM   #10
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But triangles with sides that are segments of a diagonal are not included?
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.)
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