Mock ARML problem

isabella2296

Pretty sure this is the right place to post this (as the forum description mentions ARML.)

I'd like a hint on this problem (not a solution!)

Given that the sum of all positive integers with exactly two proper divisors, each of which is less than 30, is 2397, find the sum of all positive integers with exactly three proper divisors, each of which is less than 30 (a proper divisor of n is a positive integer that divides but is not equal to n).

Thanks!

**Posted:** Thu Nov 05, 2009 9:58 am

_________________
Czech out $\mathfrak {Encyclopedia } \mathfrak { Mathematica}$
Latest entry: Nine-Point Center

darkmage2009

Is this an individual or a team round question?

Regardless, think about the formula of the number of divisors of an integer. Then you should realize what form an integer with exactly two proper divisors and an integer with exactly three proper divisors would take. Then you do a little algebraic manipulation and arithmetic to find your answer.

**Posted:** Thu Nov 05, 2009 10:24 am

_________________
1+1=2 in base 10, and 1+1=10 in base 2. Coincidence? I think not!

isabella2296

**Posted:** Thu Nov 05, 2009 5:29 pm

_________________
Czech out $\mathfrak {Encyclopedia } \mathfrak { Mathematica}$
Latest entry: Nine-Point Center

andersonw · Yang-Mills Theory Offline Joined: 23 Oct 2007 Posts: 617

I remember this was from an Altheman mock ARML. Not sure which one, though.

hint

**Posted:** Thu Nov 05, 2009 6:16 pm

10000th User

Let me state a fact that would perhaps help some of the unexperienced readers:

Let $n=p_1^{a_1}p_2^{a_2}\cdots p_j^{a_j}$ be the prime factorization of $n>0$ . Then, the number of positive divisors of $n$ is $\prod_{k=1}^j(a_k+1)=(a_1+1)(a_2+1)\cdots(a_j+1)$ .

**Posted:** Thu Nov 05, 2009 9:09 pm

_________________
$\mathrm{Hom}(\mathbb{Z},\mathbb{Z})\cong\mathbb{Z}$
Disclaimer: I'll revive old threads if: 1) something ambiguous in the solution or wrong argument, 2) no solution, 3) I got new insight to input.

isabella2296

Yay, I think I got it!

Is the answer 7282?

**Posted:** Fri Nov 06, 2009 4:33 am

_________________
Czech out $\mathfrak {Encyclopedia } \mathfrak { Mathematica}$
Latest entry: Nine-Point Center

10000th User

**Posted:** Fri Nov 06, 2009 7:05 am

_________________
$\mathrm{Hom}(\mathbb{Z},\mathbb{Z})\cong\mathbb{Z}$
Disclaimer: I'll revive old threads if: 1) something ambiguous in the solution or wrong argument, 2) no solution, 3) I got new insight to input.

isabella2296

Thanks for the help, everyone!

**Posted:** Fri Nov 06, 2009 9:42 am

_________________
Czech out $\mathfrak {Encyclopedia } \mathfrak { Mathematica}$
Latest entry: Nine-Point Center