Difference between revisions of "2002 AMC 12B Problems/Problem 12"
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=== Solution 1 === | === Solution 1 === | ||
− | Let <math>x^2 = \frac{n}{20-n} </math>, with <math>x \ge 0</math> (note that the solutions <math>x < 0</math> do not give any additional solutions for <math>n</math>). Then rewriting, <math>n = \frac{20x^2}{x^2 + 1}</math>. Since <math>\text{gcd}(x^2, x^2 + 1) = 1</math>, it follows that <math>x^2 + 1</math> divides <math>20</math>. Listing the factors of <math>20</math>, we find that <math>x = 0, 1, 2 , 3</math> are the only <math> | + | Let <math>x^2 = \frac{n}{20-n} </math>, with <math>x \ge 0</math> (note that the solutions <math>x < 0</math> do not give any additional solutions for <math>n</math>). Then rewriting, <math>n = \frac{20x^2}{x^2 + 1}</math>. Since <math>\text{gcd}(x^2, x^2 + 1) = 1</math>, it follows that <math>x^2 + 1</math> divides <math>20</math>. Listing the factors of <math>20</math>, we find that <math>x = 0, 1, 2 , 3</math> are the only <math>\boxed{\mathrm{(D)} 4}</math> solutions (respectively yielding <math>n = 0, 10, 16, 18</math>). |
=== Solution 2 === | === Solution 2 === |
Revision as of 23:30, 17 November 2013
- The following problem is from both the 2002 AMC 12B #12 and 2002 AMC 10B #16, so both problems redirect to this page.
Problem
For how many integers is
the square of an integer?
Solution
Solution 1
Let , with
(note that the solutions
do not give any additional solutions for
). Then rewriting,
. Since
, it follows that
divides
. Listing the factors of
, we find that
are the only
solutions (respectively yielding
).
Solution 2
For and
the fraction is negative, for
it is not defined, and for
it is between 0 and 1.
Thus we only need to examine and
.
For and
we obviously get the squares
and
respectively.
For prime the fraction will not be an integer, as the denominator will not contain the prime in the numerator.
This leaves , and a quick substitution shows that out of these only
and
yield a square.
See also
2002 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 | ||
All AMC 10 Problems and Solutions |
2002 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.