# Contest Question

• Apr 21st 2012, 05:34 PM
rrco
Contest Question
Taken from the final question in a recently completed math contest:

The positive divisors of 21 are 1,3,7, and 21. Let S(n) be the sum of the positive divisors of the positive integer n. For example S(21)=1+3+7+21=32.

a) If p is an odd prime integer, find the value of p such that S(2p^2) = 2613
b) The consecutive integers 14 and 15 have the property that S(14) = S(15). Determine all pairs of consecutive integers m and n such that m=2p and n=9q for prime integers p,q > 3, and S(m) = S(n).
c)Determine the number of pairs of distinct prime integers p and q, each less than 30, with the property that S(p^3q) is not divisible by 24.

I've managed to answer both a, b however for the life of me I cannot begin to think of a way to prove c. The math contest is over however the fact that I cannot answer the last question has been bugging me ever since. Help please? :D Also please provide a explanation to your solution rather than just throwing out a number if possible. As I would really like to see the though processes behind how other's approached these questions.
• Apr 24th 2012, 09:01 AM
ignite
Re: Contest Question
First note that it is easier to find number of pairs such that $\displaystyle S(p^{3q})$ is divisible by 24.
There are 10 primes less than 30.So in total we have 10*10=100 pairs of primes.
Next we find pairs with the desired property.
It is quite obvious that $\displaystyle S(p^{3q})=1+p+p^2+....+p^{3q}$.......(*)
If 24 divides the above summation then both 3 and 8 divide the summation.
$\displaystyle \Rightarrow p \ne 2,3$
Suppose $\displaystyle p \equiv 1$ (mod 3) $\displaystyle \Rightarrow (1)(3q+1) \equiv 0$ (mod 3),which is not possible.
$\displaystyle \Rightarrow p \equiv -1$ (mod 3).......(1)
This also implies $\displaystyle q \ne 2$ . We make use of this property that q is odd later.
Consider following cases:
Case I:$\displaystyle p \equiv 1$ (mod 8)
$\displaystyle \Rightarrow 1(3q+1) \equiv 0$ (mod 8) [From Equation (*)]
This along with Equation (1) gives p=17 and q=5,13,29
Case II:$\displaystyle p \equiv 3$ (mod 8)
$\displaystyle \Rightarrow 4(\frac{3q+1}{2}) \equiv 0$ (mod 8)
$\displaystyle \Rightarrow 2(3q+1) \equiv 0$ (mod 8)
This along with Equation (1) gives p=11 and q=5,13,17,29
Case III:$\displaystyle p \equiv 5$ (mod 8)
$\displaystyle \Rightarrow 6(\frac{3q+1}{2}) \equiv 0$ (mod 8)
$\displaystyle \Rightarrow 3(3q+1) \equiv 0$ (mod 8)
This along with Equation (1) gives p=5,29 and q=5,13,29
Case IV:$\displaystyle p \equiv 7$ (mod 8)
$\displaystyle \Rightarrow 0 \equiv 0$ (mod 8)
This along with equation (1) gives p=23 and q=3,5,7,11,13,17,19,23,29

Total number of pairs=22

Please note that if p and q are taken to be distinct,then answer would be (10)(9)-(19)=71.

Please notify me if you don't understand something or find something wrong.
• Apr 24th 2012, 11:50 AM
rrco
Re: Contest Question
Sorry I think I misled you with my original post. Question c) should instead read:

S(p^3*q)

My greatest apologies ignite. However I took a look at your answer, and from what I understand (which is very little ;o) you had a great solution.
• Apr 24th 2012, 12:00 PM
ignite
Re: Contest Question
Oh!Do you mean $\displaystyle S(p^3q)$?
If so,then it's quite easy by reasoning similar to my above post.
Also here is python code which will print all such pairs:
Code:

p=[2,3,5,7,11,13,17,19,23,29] for i in p:     for j in p:         if i<=j and (1+i+i**2+i**3)*(1+j)%24==0:             print i,j
There are total of 37 pairs.
• May 2nd 2012, 05:08 PM
rrco
Re: Contest Question
Hey Ignite,

Just to follow up on my original post. Waterloo recently released the answer key for the 2012 Hypatia Contest. As it turns out the answer to this question was 23. I'm still not quite sure how they got to that answer. But would you mind taking a look at the answer key and interpreting it? If you do thanks so much :D.