Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2005 AMC 10A Problems/Problem 24"
m
Line 21: Line 21:
 
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 [[factor]]s of <math>48</math>, we have several possibilities to solve for <math>p_{1}</math> and <math>p_{2}</math>:
+
Looking at pairs of [[divisor]]s of <math>48</math>, we have several possibilities to solve for <math>p_{1}</math> and <math>p_{2}</math>:
  
  

Revision as of 08:44, 11 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 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_24&oldid=9404"