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Difference between revisions of "2005 AMC 10A Problems/Problem 24"

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<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 5 </math>
 
<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 5 </math>
  
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==Solution==
+
==Solution 1==
 +
If <math> P(n) = \sqrt{n} </math>, then <math> n = p_{1}^{2} </math>, where <math> p_{1} </math> is a [[prime number]].
 +
 
 +
If <math> P(n+48) = \sqrt{n+48} </math>, then <math>n + 48</math> is a square, but we know that n is <math>p_{1}^{2} </math>.
 +
 
 +
 
 +
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.
 +
 
 +
<math>4 , 9 , 25 , 49 , 121, 169 , 289 , 361 , 529</math>
 +
 
 +
 
 +
But adding 48 to a number ending with 9 will result in a number ending with 7 and we know that a perfect square does not end in 7 hence we can eliminate those cases to get the new list.
 +
 
 +
<math>4 , 25 , 49 , 121 , 361</math>
 +
 
 +
 
 +
Substituting the value's we get 121 as the only possible solution.
 +
Hence the answer is (B).
 +
 
 +
~<i>rudolf1279<i>
 +
 
 +
==Solution 2==
 
If <math> P(n) = \sqrt{n} </math>, then <math> n = p_{1}^{2} </math>, where <math> p_{1} </math> is a [[prime number]].  
 
If <math> P(n) = \sqrt{n} </math>, then <math> n = p_{1}^{2} </math>, where <math> p_{1} </math> is a [[prime number]].  
  

Revision as of 20:01, 7 October 2020

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 and we know that a perfect square does not end in 7 hence we can eliminate those cases to get the new list.

$4 , 25 , 49 , 121 , 361$


Substituting the value's we get 121 as the only possible solution. Hence the answer is (B).

~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

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