Difference between revisions of "2007 AMC 12B Problems/Problem 24"
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<math>mn = 126</math>. | <math>mn = 126</math>. | ||
− | Factoring 126, we get <math>6</math> pairs of numbers: <math>(1,126), (2,63), (3,42), (6,21), (7,18) | + | Factoring 126, we get <math>6</math> pairs of numbers: <math>(1,126), (2,63), (3,42), (6,21), (7,18),</math> and <math>(9,14)</math>. |
− | Looking back at our equations for <math>m</math> and <math>n</math>, we can solve for <math>k = \frac{2m + 2n}{18} = \frac{m+n}{9}</math>. Since <math>k</math> is an integer, there are only <math>2</math> pairs of <math>(m,n)</math> that work: <math>(3,42) | + | Looking back at our equations for <math>m</math> and <math>n</math>, we can solve for <math>k = \frac{2m + 2n}{18} = \frac{m+n}{9}</math>. Since <math>k</math> is an integer, there are only <math>2</math> pairs of <math>(m,n)</math> that work: <math>(3,42)</math> and <math>(6,21)</math>. This means that there are <math>2</math> values of <math>k</math> such that <math>u</math> is an integer. But looking back at <math>u</math> in terms of <math>k</math>, we have <math>\pm</math>, meaning that there are <math>2</math> values of <math>u</math> for every <math>k</math>. Thus, the answer is <math>2 \cdot 2 = 4 \Rightarrow \mathrm{(A)}</math>. |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2007|ab=B|num-b=23|num-a=25}} | {{AMC12 box|year=2007|ab=B|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:19, 2 May 2015
Problem 24
How many pairs of positive integers are there such that and is an integer?
Solution
Combining the fraction, must be an integer.
Since the denominator contains a factor of ,
Since for some positive integer , we can rewrite the fraction(divide by on both top and bottom) as
Since the denominator now contains a factor of , we get .
But since , we must have , and thus .
For the original fraction simplifies to .
For that to be an integer, must divide , and therefore we must have . Each of these values does indeed yield an integer.
Thus there are four solutions: , , , and the answer is
Solution 2
Let's assume that We get--
Factoring this, we get equations-
(It's all negative, because if we had positive signs, would be the opposite sign of )
Now we look at these, and see that-
This gives us solutions, but we note that the middle term needs to give you back .
For example, in the case
, the middle term is , which is not equal by for whatever integar .
Similar reason for the fourth equation. This elimnates the last four solutions out of the above eight listed, giving us 4 solutions total
Solution 3
Let . Then the given equation becomes .
Let's set this equal to some value, .
Clearing the denominator and simplifying, we get a quadratic in terms of :
Since and are integers, is a rational number. This means that is an integer.
Let . Squaring and rearranging yields:
.
In order for both and to be an integer, and must both be odd or even. (This is easily proven using modular arithmetic.) In the case of this problem, both must be even. Let and .
Then:
.
Factoring 126, we get pairs of numbers: and .
Looking back at our equations for and , we can solve for . Since is an integer, there are only pairs of that work: and . This means that there are values of such that is an integer. But looking back at in terms of , we have , meaning that there are values of for every . Thus, the answer is .
See Also
2007 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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.