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Revision as of 12:01, 26 November 2015

Problem

For each positive integer $n > 1$, let $P(n)$ denote the greatest prime factor of $n$. 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 $p_{2}$:


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

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

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

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


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

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

$p_{1} = 11$

$p_{2} = 13$


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

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

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

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


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

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

$p_{1} = 4$

$p_{2} = 8$


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

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

$p_{1} = 1$

$p_{2} = 7$


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 • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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
All AMC 10 Problems and Solutions

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