Difference between revisions of "2002 AMC 12B Problems/Problem 12"
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Simon's Favourite Factoring Trick | Simon's Favourite Factoring Trick | ||
Since \frac{n}{20-n} is an integer k, w multiple both sides by 20-n. This gives us <math>n</math> = <math>20k^2</math>-<math>nk^2</math>. We subtract <math>20k^2</math> on both sides, then add <math>nk^2</math> on both sides as a prerequisite for using Simon's Favorite Factoring Trick. We have (k^2+1)(n-20)=20. We then consider the different factors of 20 that k^2+1 can be. It could be <math>1</math>,<math>2</math>,<math>4</math>,<math>5</math>,<math>10</math>, and <math>20</math>. After checking case by case, we then are able to identify that there are 4 such k values that also yield an integer n value, meaning that there are 4 values, so the correct answer is <math>\boxed{(D)}</math> | Since \frac{n}{20-n} is an integer k, w multiple both sides by 20-n. This gives us <math>n</math> = <math>20k^2</math>-<math>nk^2</math>. We subtract <math>20k^2</math> on both sides, then add <math>nk^2</math> on both sides as a prerequisite for using Simon's Favorite Factoring Trick. We have (k^2+1)(n-20)=20. We then consider the different factors of 20 that k^2+1 can be. It could be <math>1</math>,<math>2</math>,<math>4</math>,<math>5</math>,<math>10</math>, and <math>20</math>. After checking case by case, we then are able to identify that there are 4 such k values that also yield an integer n value, meaning that there are 4 values, so the correct answer is <math>\boxed{(D)}</math> | ||
+ | ~CharmaineMa07292010 | ||
== See also == | == See also == |
Revision as of 05:54, 17 October 2023
- 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 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. Therefore, there are only
solutions (respectively yielding
).
Solution 3
If , then
and
, otherwise
will be negative. Thus
and
Checking all
for which
, we have
,
,
,
as the possibilities.
~ Nafer
Solution 4
For all integers x, is always a positive integer. So solve for
, getting
and
, getting
. For all values n less than 0 and greater than 20, the value
is negative, so now try values of n between 10 and 20. Quick substitution finds
,
,
, and
which yields
,
,
, and
respectively. 4 values, or
Solution 5
Simon's Favourite Factoring Trick
Since \frac{n}{20-n} is an integer k, w multiple both sides by 20-n. This gives us =
-
. We subtract
on both sides, then add
on both sides as a prerequisite for using Simon's Favorite Factoring Trick. We have (k^2+1)(n-20)=20. We then consider the different factors of 20 that k^2+1 can be. It could be
,
,
,
,
, and
. After checking case by case, we then are able to identify that there are 4 such k values that also yield an integer n value, meaning that there are 4 values, so the correct answer is
~CharmaineMa07292010
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.