1995 AIME Problems/Problem 6

Problem

Let $n=2^{31}3^{19}.$ How many positive integer divisors of $n^2$ are less than $n_{}$ but do not divide $n_{}$ ?

Solution 1

We know that $n^2 = 2^{62}3^{38}$ must have $(62+1)\times (38+1)$ factors by its prime factorization. If we group all of these factors (excluding $n$ ) into pairs that multiply to $n^2$ , then one factor per pair is less than $n$ , and so there are $\frac{63\times 39-1}{2} = 1228$ factors of $n^2$ that are less than $n$ . There are $32\times20-1 = 639$ factors of $n$ , which clearly are less than $n$ , but are still factors of $n^2$ . Therefore, there are $1228-639=\boxed{589}$ factors of $n$ that do not divide $n^2$ .

Solution 2

Let $n=p_1^{k_1}p_2^{k_2}$ for some prime $p_1,p_2$ . Then $n^2$ has $\frac{(2k_1+1)(2k_2+1)-1}{2}$ factors less than $n$ .

This simplifies to $\frac{4k_1k_2+2k_1+2k_2}{2}=2k_1k_2+k_1+k_2$ .

The number of factors of $n$ less than $n$ is equal to $(k_1+1)(k_2+1)-1=k_1k_2+k_1+k_2$ .

Thus, our general formula for $n=p_1^{k_1}p_2^{k_2}$ is

$\text{Number\;of\;factors\;that\;satisfy\;conditions\;}=(2k_1k_2+k_1+k_2)-(k_1k_2+k_1+k_2)=k_1k_2<math>.$

Incorporating this into our problem gives </math>19\times31=\boxed{589}$.

1995 AIME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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1995 AIME Problems/Problem 6

Contents

Problem

Solution 1

Solution 2

See also