2007 AMC 10B Problems/Problem 25
Contents
Problem
How many pairs of positive integers are there such that
and
have no common factors greater than
and:
is an integer?
Solutions
Solution 1
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 numerator and denominator) 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 be a factor of
, 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 any integer
.
Similar reason for the fourth equation. This eliminates 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
.
Solution 4
Rewriting the expression over a common denominator yields . This expression must be equal to some integer
.
Thus, . Taking this
yields
. Since
,
. This implies that
so
.
We can then take to get that
. Thus
.
However, taking ,
so
cannot equal 1.
Also, note that if ,
. Since
,
will be an integer, but
will not be an integer since none of the possible values of
are multiples of 9. Thus,
cannot equal 9.
Thus, the only possible values of is 3, and
can be 1, 2, 7, or 14. This yields 4 possible solutions, so the answer is
.
Solution 5 (Similar to Solution 1)
Rewriting over a common denominator gives
Thus, we have
Next, we have
Thus,
Next, we have
Thus,
Now, we simply do casework on
Plugging in and
gives that there are
total solutions for
~coolmath2017
Solution 6 (Similar to solution 3)
Let So
, where I is an integer. Algebraic manipulations yield:
. The discriminant of this must be the square of a rational number, call this R. So
. I is
the sum of
and
. To have an integer sum,
and
must have the same denominator, namely 3. We proceed with casework.
Case 1.
,
. This yields
, which is not an integer. This case produces 0 solutions.
Case 2.
,
. This yields
. Substituting into our original equation yields:
. Factoring gives:
,
. This case produces 2 solutions, namely (1,3) and (14,3).
Case 3.
,
. This yields
. Substituting into our original equation yields:
. Factoring gives:
,
. This case produces 2 solutions, namely (2,3) and (7,3).
Case 4.
,
. This yields
, which is not an integer. This case produces 0 solutions.
Altogether, we have 4 solutions, so our answer is .
~Math4Life2020
See Also
2007 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last question | |
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