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Difference between revisions of "1987 AHSME Problems/Problem 23"

(Created page with "==Problem== If <math>p</math> is a prime and both roots of <math>x^2+px-444p=0</math> are integers, then <math>\textbf{(A)}\ 1<p\le 11 \qquad \textbf{(B)}\ 11<p \le 21 \qquad \...")
 
(Added a solution with explanation)
 
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\textbf{(E)}\ 41< p\le 51</math>
 
\textbf{(E)}\ 41< p\le 51</math>
 
    
 
    
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== Solution ==
 +
For integer roots, we need the discriminant, which is <math>p^2 - 4 \cdot 1 \cdot (-444p) = p^2 + 1776p = p(p+1776)</math>, to be a perfect square. Now, this means that <math>p</math> must divide <math>p + 1776</math>, as if it did not, there would be a lone prime factor of <math>p</math>, and so this expression could not possibly be a perfect square. Thus <math>p</math> divides <math>p + 1776</math>, which implies <math>p</math> divides <math>1776 = 2^{4} \cdot 3 \cdot 37</math>, so we must have <math>p = 2</math>, <math>3</math>, or <math>37</math>. It is easy to verify that neither <math>p = 2</math> nor <math>p = 3</math> make <math>p(p+1776)</math> a perfect square, but <math>p = 37</math> does, so the answer is <math>31 < p \le 41</math>, which is <math>\boxed{\text{D}}</math>.
  
 
== See also ==
 
== See also ==
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{{AHSME box|year=1987|num-b=21|num-a=23}}   
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{{AHSME box|year=1987|num-b=22|num-a=24}}   
  
 
[[Category: Intermediate Algebra Problems]]
 
[[Category: Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:57, 1 March 2018

Problem

If $p$ is a prime and both roots of $x^2+px-444p=0$ are integers, then

$\textbf{(A)}\ 1<p\le 11 \qquad \textbf{(B)}\ 11<p \le 21 \qquad \textbf{(C)}\ 21< p \le 31 \\ \qquad \textbf{(D)}\ 31< p\le 41\qquad \textbf{(E)}\ 41< p\le 51$

Solution

For integer roots, we need the discriminant, which is $p^2 - 4 \cdot 1 \cdot (-444p) = p^2 + 1776p = p(p+1776)$, to be a perfect square. Now, this means that $p$ must divide $p + 1776$, as if it did not, there would be a lone prime factor of $p$, and so this expression could not possibly be a perfect square. Thus $p$ divides $p + 1776$, which implies $p$ divides $1776 = 2^{4} \cdot 3 \cdot 37$, so we must have $p = 2$, $3$, or $37$. It is easy to verify that neither $p = 2$ nor $p = 3$ make $p(p+1776)$ a perfect square, but $p = 37$ does, so the answer is $31 < p \le 41$, which is $\boxed{\text{D}}$.

See also

1987 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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