2003 AMC 12B Problems/Problem 18
Contents
Problem
Let and
be positive integers such that
The minimum possible value of
has a prime factorization
What is
Solution 1
Substitute into
. We then have
. Divide both sides by
, and it follows that:
Note that because and
are prime, the minimum value of
must involve factors of
and
only. Thus, we try to look for the lowest power
of
such that
, so that we can take
to the fifth root. Similarly, we want to look for the lowest power
of
such that
. Again, this allows us to take the fifth root of
. Obviously, we want to add
to
and subtract
from
because
and
are multiplied by
and divided by
, respectively. With these conditions satisfied, we can simply multiply
and
and substitute this quantity into
to attain our answer.
We can simply look for suitable values for and
. We find that the lowest
, in this case, would be
because
. Moreover, the lowest
should be
because
. Hence, we can substitute the quantity
into
. Doing so gets us:
Taking the fifth root of both sides, we are left with .
Solution 2
A simpler way to tackle this problem without all that modding is to keep the equation as:
As stated above, and
must be the factors 7 and 11 in order to keep
at a minimum. Moving all the non-y terms to the left hand side of the equation, we end up with:
The above equation means that must also contain only the factors 7 and 11 (again, in order to keep
at a minimum), so we end up with:
( and
are arbitrary variables placed in order to show that
could have more than just one 7 or one 11 as factors)
Since 7 and 11 are prime, we know that and
. The smallest positive combinations that would work are
and
. Therefore,
.
is correct.
See Also
2003 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 |
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