2016 AMC 12B Problems/Problem 24
Problem
There are exactly ordered quadruplets
such that
and
. What is the smallest possible value for
?
Solution
Let , etc., so that
. Then for each prime power
in the prime factorization of
, at least one of the prime factorizations of
has
, at least one has
, and all must have
with
.
Let be the number of ordered quadruplets of integers
such that
for all
, the largest is
, and the smallest is
. Then for the prime factorization
we must have
So let's take a look at the function
by counting the quadruplets we just mentioned..
There are quadruplets which consist only of
and
. Then there are
quadruplets which include three different values, and
with four. Thus
and the first few values from
onwards are
Straight away we notice that
, so the prime factorization of
can use the exponents
. To make it as small as possible, assign the larger exponents to smaller primes. The result is
, so
which is answer
.
See Also
2016 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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All AMC 12 Problems and Solutions |