2010 AMC 12B Problems/Problem 9
Problem 9
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
A number whose square root is a perfect cube and whose cube root is a perfect square will be in the form where a is an integer, because
is the
of
and
. A number that is divisible by
obviously ends in a
. The only way
can end in a zero is if
ends in a zero. The smallest number that ends in a
is
, so our number
, with
digits.
See also
2010 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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