2010 AMC 12B Problems/Problem 25
Problem 25
For every integer , let
be the largest power of the largest prime that divides
. For example
. What is the largest integer
such that
divides
?
Solution
Because 67 is the largest prime factor of 2010, it means that in the prime factorization of , there'll be
where
is the desired value we are looking for. Thus, to find this answer, we need to look for the number of times 67 would be incorporated into the giant product.
Any number of the form
would fit this form. However, this number tops at
because 71 is a higher prime than 67.
itself must be counted twice because it's counted twice as a squared number. Any non-prime number that's less than 79 (and greater than 71) can be counted, and this totals 5. We have
numbers (as 71 isn't counted - 1 through 70), an additional
(
), and
values just greater than
but less than
(72, 74, 75, 76, 77, and 78). Thus,
Similar Solution
After finding the prime factorization of $2010=2\cdot3\cdot5\67$ (Error compiling LaTeX. Unknown error_msg), divide by
and add
divided by
in order to find the total number of multiples of
between
and
.
Since
,
, and
are prime numbers greater than
and less than or equal to
, subtract
from
to get the answer
.
See Also
2010 AMC 12B (Problems • Answer Key • Resources) | |
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