Difference between revisions of "2006 AIME A Problems/Problem 3"
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<math>\displaystyle \lfloor \frac{200}{3}\rfloor+\lfloor\frac{200}{9}\rfloor+\lfloor \frac{200}{27}\rfloor+\lfloor\frac{200}{81}\rfloor =66+22+7+2=97</math> | <math>\displaystyle \lfloor \frac{200}{3}\rfloor+\lfloor\frac{200}{9}\rfloor+\lfloor \frac{200}{27}\rfloor+\lfloor\frac{200}{81}\rfloor =66+22+7+2=97</math> | ||
+ | |||
threes in <math>\displaystyle 200!</math> and | threes in <math>\displaystyle 200!</math> and | ||
<math>\displaystyle \lfloor \frac{100}{3}\rfloor+\lfloor\frac{100}{9}\rfloor+\lfloor \frac{100}{27}\rfloor+\lfloor\frac{100}{81}\rfloor=33+11+3+1=48 </math> | <math>\displaystyle \lfloor \frac{100}{3}\rfloor+\lfloor\frac{100}{9}\rfloor+\lfloor \frac{100}{27}\rfloor+\lfloor\frac{100}{81}\rfloor=33+11+3+1=48 </math> |
Revision as of 14:29, 24 July 2006
Problem
Let be the product of the first
positive odd integers. Find the largest integer
such that
is divisible by
Solution
Note that the product of the first positive odd integers can be written as
Hence, we seek the number of threes in decreased by the number of threes in
There are
threes in and
threes in
Therefore, we have a total of threes.