Difference between revisions of "1987 AHSME Problems/Problem 23"
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\textbf{(E)}\ 41< p\le 51</math> | \textbf{(E)}\ 41< p\le 51</math> | ||
| − | + | == 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 == | ||
Latest revision as of 13:57, 1 March 2018
Problem
If 
is a prime and both roots of 
are integers, then
Solution
For integer roots, we need the discriminant, which is 
, to be a perfect square. Now, this means that must divide 
, as if it did not, there would be a lone prime factor of , and so this expression could not possibly be a perfect square. Thus divides , which implies divides 
, so we must have 
, 
, or 
. It is easy to verify that neither nor 
make 
a perfect square, but 
does, so the answer is 
, which is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
