2009 AMC 12A Problems/Problem 18
- The following problem is from both the 2009 AMC 12A #18 and 2009 AMC 10A #25, so both problems redirect to this page.
Problem
For , let
, where there are
zeros between the
and the
. Let
be the number of factors of
in the prime factorization of
. What is the maximum value of
?
Solution
The number can be written as
.
For we have
. The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have
.
For we have
. For
the value in the parentheses is odd, hence
.
This leaves the case . We have
. The value
is obviously even. And as
, we have
, and therefore
. Hence the largest power of
that divides
is
, and this gives us the desired maximum of the function
:
.
Alternate Solution
Notice that 2 is a prime factor of if and only if
is even. Also notice that regardless of the positive integral value of
, dividing by
leaves us with a terminal digit of zero. Dividing by 2 again makes
odd, leaving us with the maximum value of
See Also
2009 AMC 12A (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 |
2009 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Question | |
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 10 Problems and Solutions |