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2005 AMC 10A Problems/Problem 24

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Problem

For each positive integer $m > 1$ , let $P(m)$ denote the greatest prime factor of $m$ . For how many positive integers $n$ is it true that both $P(n) = \sqrt{n}$ and $P(n+48) = \sqrt{n+48}$ ?

$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 5$

Solution

If $P(n) = \sqrt{n}$ , then $n = p_{1}^{2}$ , where $p_{1}$ is a prime number.

If $P(n+48) = \sqrt{n+48}$ , then $n+48 = p_{2}^{2}$ , where $p_{2}$ is a different prime number.

So:

$p_{2}^{2} = n+48$

$p_{1}^{2} = n$

$p_{2}^{2} - p_{1}^{2} = 48$

$(p_{2}+p_{1})(p_{2}-p_{1})=48$

Since $p_{1} > 0$ : $(p_{2}+p_{1}) > (p_{2}-p_{1})$ .

Looking at pairs of divisors of $48$ , we have several possibilities to solve for $p_{1}$ and $p_{2}$ :

$(p_{2}+p_{1}) = 48$

$(p_{2}-p_{1}) = 1$

$p_{1} = \frac{47}{2}$

$p_{2} = \frac{49}{2}$

$(p_{2}+p_{1}) = 24$

$(p_{2}-p_{1}) = 2$

$(p_{2}+p_{1}) = 16$

$(p_{2}-p_{1}) = 3$

$p_{1} = \frac{13}{2}$

$p_{2} = \frac{19}{2}$

$(p_{2}+p_{1}) = 12$

$(p_{2}-p_{1}) = 4$

$(p_{2}+p_{1}) = 8$

$(p_{2}-p_{1}) = 6$

The only solution $(p_{1} , p_{2})$ where both numbers are primes is $(11,13)$ .

Therefore the number of positive integers $n$ that satisfy both statements is $1\Rightarrow \mathrm{(B)}$

See Also

2005 AMC 10A Problems

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