Difference between revisions of "2006 AIME II Problems/Problem 3"
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For more information, see also [[Factorial#Prime factorization| prime factorizations of a factorial]]. | For more information, see also [[Factorial#Prime factorization| prime factorizations of a factorial]]. | ||
+ | |||
+ | == Solution Two == | ||
+ | We count the multiples of <math>3^k</math> below 200 and subtract the count of multiples of <math>2\cdot 3^k</math>: | ||
+ | |||
+ | <cmath>\left\lfloor \frac{200}{3}\right\rfloor - \left\lfloor \frac{200}{6}\right\rfloor +\left\lfloor\frac{200}{9}\right\rfloor - \left\lfloor \frac{200}{18}\right\rfloor +\left\lfloor \frac{200}{27}\right\rfloor - \left\lfloor \frac{200}{54}\right\rfloor+\left\lfloor\frac{200}{81}\right\rfloor - \left\lfloor \frac{200}{162}\right\rfloor</cmath> | ||
+ | <cmath>= 66 - 33 + 22 - 11 + 7 - 3 + 2 - 1 = 49.</cmath> | ||
== See also == | == See also == | ||
* [[Number Theory]] | * [[Number Theory]] |
Revision as of 02:33, 2 July 2014
Contents
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.
For more information, see also prime factorizations of a factorial.
Solution Two
We count the multiples of below 200 and subtract the count of multiples of :
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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