1995 AIME Problems/Problem 6

Problem

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

Solution

We know that $n^2$ must have $63\times 39$ factors by its prime factorization. There are $\frac{63\times 39-1}{2} = 1228$ factors of $n^2$ that are less than $n$ , because if they form pairs $a$ , then there is one factor per pair that is less than $n$ . There are $32\times20-1 = 639$ factors of $n$ that are less than $n$ itself. These are also factors of $n^2$ . Therefore, there are $1228-639=539$ factors of $n$ that do not divide $n$ .

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

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