1996 AHSME Problems/Problem 29

Problem

If $n$ is a positive integer such that $2n$ has $28$ positive divisors and $3n$ has $30$ positive divisors, then how many positive divisors does $6n$ have?

$\text{(A)}\ 32\qquad\text{(B)}\ 34\qquad\text{(C)}\ 35\qquad\text{(D)}\ 36\qquad\text{(E)}\ 38$

Solution

Working with the second part of the problem first, we know that $3n$ has $30$ divisors. We try to find the various possible prime factorizations of $3n$ by splitting $30$ into various products of $1, 2$ or $3$ integers.

$30 \rightarrow p^{29}$

$2 \cdot 15 \rightarrow pq^{14}$

$3\cdot 10 \rightarrow p^2q^9$

$5\cdot 6 \rightrarrow p^4q^5$ (Error compiling LaTeX. Unknown error_msg)

$2\cdot 3\cdot 5 \rightarrow pq^2r^4$

The variables $p, q, r$ are different prime factors, and one of them must be $3$ . We now try to count the factors of $2n$ , to see which prime factorization is correct and has $28$ factors.

In the first case, $p=3$ is the only possibility. This gives $2n = 2\cdot p^{28}$ , which has $2\cdot {29}$ factors, which is way too many.

In the second case, $p=3$ gives $2n = 2q^{14}$ . If $q=2$ , then there are $16$ factors, while if $q\neq 2$ , there are $2\cdot 15 = 30$ factors.

In the second case, $q=3$ gives $2n = 2p3^{13}$ . If $p=2$ , then there are $3\cdot 13$ factors, while if $p\neq 2$ , there are $2\cdot 2 \cdot 13$ factors.

In the third case, $p=3$ gives $2n = 2\cdot 3\cdot q^9$ . If $q=2$ , then there are $11\cdot 2 = 22$ factors, while if $q \neq 2$ , there are $2\cdot 2\cdot 10$ factors.

In the third case, $q=3$ gives $2n = 2\cdot p^2\cdot 3^8$ . If $p=2$ , then there are $4\cdot 9$ factors, while if $p \neq 2$ , there are 2\cdot 3\cdot 9$factors.

In the fourth case,$ (Error compiling LaTeX. Unknown error_msg)p=3 $gives$ 2n = 2\cdot 3^3\cdot q^5 $. If$ q=2 $, then there ar$ 7\cdot 4= 28$factors. This is the factorization we want.

Thus,$ (Error compiling LaTeX. Unknown error_msg)3n = 3^4 \cdot 2^5 $, which has$ 5\cdot 6 = 30 $factors, and$ 2n = 3^3 \cdot 2^6 $, which has$ 4\cdot 7 = 28$factors.

In this case,$ (Error compiling LaTeX. Unknown error_msg)6n = 3^4\cdot 2^6 $, which has$ 5\cdot 7 = 35 $factors, and the answer is$ \boxed{C}$

1996 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 28	Followed by Problem 30
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1996 AHSME Problems/Problem 29

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