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Difference between revisions of "1956 AHSME Problems/Problem 48"

(Created page with "== Problem 48== If <math>p</math> is a positive integer, then <math>\frac {3p + 25}{2p - 5}</math> can be a positive integer, if and only if <math>p</math> is: <math> \tex...")
 
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<cmath>\frac{3p+25}{2p-5} = 1 + \frac{p+30}{2p-5}</cmath>
 
<cmath>\frac{3p+25}{2p-5} = 1 + \frac{p+30}{2p-5}</cmath>
  
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Therefore, in order for <math>\frac{3p+25}{2p-5}</math> to be a positive integer, <math>\frac{p+30}{2p-5}</math> must be a non-negative integer. Since the bottom the the fraction is an odd number, we can multiply the top of <math>\frac{p+30}{2p-5}</math> by 2 without changing whether it is an integer or not. Therefore, in order for <math>\frac{3p+25}{2p-5}</math> to be an integer, <math>\frac{2p+60}{2p-5} = 1 + \frac{65}{2p-5}</math> must also be an integer. As a result, <math>2p-5</math> must be a factor of 65, or <math>p = 2, 5, 9, 35</math>. Therefore <math>p</math> must be at least 3, and less than or equal to 35. So the answer which best fits these constraints is <math>\boxed{\textbf{(B)}}</math>.
+
Therefore, in order for <math>\frac{3p+25}{2p-5}</math> to be a positive integer, <math>\frac{p+30}{2p-5}</math> must be a non-negative integer. Since the bottom the the fraction is an odd number, we can multiply the top of <math>\frac{p+30}{2p-5}</math> by 2 without changing whether it is an integer or not. Therefore, in order for <math>\frac{3p+25}{2p-5}</math> to be an integer, <math>\frac{2p+60}{2p-5} = 1 + \frac{65}{2p-5}</math> must also be an integer. As a result, <math>2p-5</math> must be a factor of 65, or <math>p = 3, 5, 9, 35</math>. Therefore <math>p</math> must be at least 3, and less than or equal to 35. So the answer which best fits these constraints is <math>\boxed{\textbf{(B)}}</math>.

Latest revision as of 05:05, 13 January 2024

Problem 48

If $p$ is a positive integer, then $\frac {3p + 25}{2p - 5}$ can be a positive integer, if and only if $p$ is:

$\textbf{(A)}\ \text{at least }3\qquad \textbf{(B)}\ \text{at least }3\text{ and no more than }35\qquad \\ \textbf{(C)}\ \text{no more than }35 \qquad \textbf{(D)}\ \text{equal to }35 \qquad \textbf{(E)}\ \text{equal to }3\text{ or }35$

Solution

Lets begin by noticing that: \[\frac{3p+25}{2p-5} = 1 + \frac{p+30}{2p-5}\]

Therefore, in order for $\frac{3p+25}{2p-5}$ to be a positive integer, $\frac{p+30}{2p-5}$ must be a non-negative integer. Since the bottom the the fraction is an odd number, we can multiply the top of $\frac{p+30}{2p-5}$ by 2 without changing whether it is an integer or not. Therefore, in order for $\frac{3p+25}{2p-5}$ to be an integer, $\frac{2p+60}{2p-5} = 1 + \frac{65}{2p-5}$ must also be an integer. As a result, $2p-5$ must be a factor of 65, or $p = 3, 5, 9, 35$. Therefore $p$ must be at least 3, and less than or equal to 35. So the answer which best fits these constraints is $\boxed{\textbf{(B)}}$.

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