June 18th, 2014, 07:54 PM  #1 
Senior Member Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53  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 $m\,(\text{mod}\, 2014)$.

June 18th, 2014, 09:09 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,354 Thanks: 2085 Math Focus: Mainly analysis and algebra 
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 
June 18th, 2014, 09:24 PM  #3  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 931 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
$$ \prod_{p\le2014}p^e+1 $$ where the product is over primes and the exponent e is the padic 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 5782digit 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  
June 19th, 2014, 04:26 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,354 Thanks: 2085 Math Focus: Mainly analysis and algebra 
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).

June 19th, 2014, 12:18 PM  #5 
Senior Member Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53  
June 20th, 2014, 05:29 AM  #6 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 931 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  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? 
June 20th, 2014, 05:40 AM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,354 Thanks: 2085 Math Focus: Mainly analysis and algebra 
Does this include joining two triangles to make another? Or indeed, joining shapes of any type to make triangles?

June 20th, 2014, 05:47 AM  #8 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 931 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
Any triangle as long as all three sides are sides or diagonals of the regular hexagon.

June 20th, 2014, 05:50 AM  #9 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,354 Thanks: 2085 Math Focus: Mainly analysis and algebra 
But triangles with sides that are segments of a diagonal are not included?

June 20th, 2014, 06:05 AM  #10 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 931 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  

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