2010 AMC 12B Problems/Problem 9
Problem
Let be the smallest positive integer such that
is divisible by
,
is a perfect cube, and
is a perfect square. What is the number of digits of
?
Solution
We know that and
. Cubing and squaring the equalities respectively gives
. Let
. Now we know
must be a perfect
-th power because
, which means that
must be a perfect
-th power. The smallest number whose sixth power is a multiple of
is
, because the only prime factors of
are
and
, and
. Therefore our is equal to number
, with
digits
.
See also
2010 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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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