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

(added problem and solution)
 
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==Problem==
 
==Problem==
−
For each positive integer <math> m > 1 </math>, let <math>P(m)</math> denote the greatest prime factor om <math>m</math>. For how many positive integers <math>n</math> is it true that both <math> P(n) = \sqrt{n} </math> and <math> P(n+48) = \sqrt{n+48} </math>?
+
For each positive integer <math> m > 1 </math>, let <math>P(m)</math> denote the greatest prime factor of <math>m</math>. For how many positive integers <math>n</math> is it true that both <math> P(n) = \sqrt{n} </math> and <math> P(n+48) = \sqrt{n+48} </math>?
  
 
<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>
  
 
==Solution==
 
==Solution==
−
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]].  
  
 
If <math> P(n+48) = \sqrt{n+48} </math>, then <math> n+48 = p_{2}^{2} </math>, where <math> p_{2} </math> is a different prime number.  
 
If <math> P(n+48) = \sqrt{n+48} </math>, then <math> n+48 = p_{2}^{2} </math>, where <math> p_{2} </math> is a different prime number.  
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Since <math> p_{1} > 0 </math>: <math> (p_{2}+p_{1}) > (p_{2}-p_{1}) </math>.  
 
Since <math> p_{1} > 0 </math>: <math> (p_{2}+p_{1}) > (p_{2}-p_{1}) </math>.  
  
−
Looking at pairs of factors of <math>48</math>, and solving for <math>p_{1}</math> and <math>p_{2}</math>:
+
Looking at pairs of [[factor]]s of <math>48</math>, we have several possibilities to solve for <math>p_{1}</math> and <math>p_{2}</math>:
  
  
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The only solution <math> (p_{1} , p_{2}) </math> where both numbers are primes is <math>(11,13)</math>.  
 
The only solution <math> (p_{1} , p_{2}) </math> where both numbers are primes is <math>(11,13)</math>.  
  
−
Therefore the number of positive integers <math>n</math> that satisfy both statements is <math>1\Rightarrow B</math>
+
Therefore the number of [[positive integer]]s <math>n</math> that satisfy both statements is <math>1\Rightarrow \mathrm{(B)}</math>
  
 
==See Also==
 
==See Also==
 
*[[2005 AMC 10A Problems]]
 
*[[2005 AMC 10A Problems]]
 +
 +
[[Category:Introductory Number Theory Problems]]
  
 
*[[2005 AMC 10A Problems/Problem 23|Previous Problem]]
 
*[[2005 AMC 10A Problems/Problem 23|Previous Problem]]
  
 
*[[2005 AMC 10A Problems/Problem 25|Next Problem]]
 
*[[2005 AMC 10A Problems/Problem 25|Next Problem]]

Revision as of 17:15, 4 August 2006

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 factors 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

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