2005 AMC 10A Problems/Problem 24

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 1

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 $n + 48$ is a square, but we know that n is $p_{1}^{2}$ .

This means we just have to check for squares of primes, add 48 and look whether the root is a prime number. We can easily see that the difference between two consecutive square after 576 is greater than or equal to 49, Hence we have to consider only the prime numbers till 23.

Squaring prime numbers below 23 including 23 we get the following list.

$4 , 9 , 25 , 49 , 121, 169 , 289 , 361 , 529$

But adding 48 to a number ending with 9 will result in a number ending with 7, but we know that a perfect square does not end in 7, so we can eliminate those cases to get the new list.

$4 , 25 , 121 , 361$

Adding 48, we get 121 as the only possible solution. Hence the answer is (B).

The only positive integer that satisfies both requirements is 11.

edited by mobius247

Note: Solution 1

Since all primes greater than 2 are odd, we know that the difference between the squares of any two consecutive primes greater than 2 is at least $(p+2)^2-p^2=4p+4$ , where p is the smaller of the consecutive primes. For $p>11$ , $4p+4>48$ . This means that the difference between the squares of any two consecutive primes both greater than 11 is greater than 48, so $n$ and $n+48$ can't both be the squares of primes if $n=p^2$ and p>11. So, we only need to check $n=2^2, 3^2, 5^2, 7^2, 11^2$ . ~apsid

Video Solution

CHECK OUT Video Solution:https://youtu.be/IsqrsMkR-mA

~rudolf1279

Solution 2
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
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2005 AMC 10A Problems/Problem 24

Contents

Problem

Solution 1

Note: Solution 1

Video Solution

Solution 2

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