Problem:
Let P be the product of the first 100 positive odd integers. Find the largest integer k such that P is divisible by 3k.
Solution:
Of the first 100 positive odd integers, 1,3,5,…,199,
33 of them, namely 3,9,15,…,195=3(2⋅33−1), are divisible by 3;
11 of them, namely 9,27,45,…,189=9(2⋅11−1), are divisible by 9;
4 of them, namely 27,81,135,189=27(2⋅4−1), are divisible by 27; and
1 of them, namely 81, is divisible by 81.Therefore k=33+11+4+1=49
OR
Note that
P=2⋅4⋅⋯⋅200200!=2100⋅100!200!
The number of factors of 3 in the numerator is
⌊200/3⌋+⌊200/32⌋+⌊200/33⌋+⌊200/34⌋=66+22+7+2=97,
and the number of factors of 3 in the denominator is
⌊100/3⌋+⌊100/32⌋+⌊100/33⌋+⌊100/34⌋=33+11+3+1=48.
Therefore k=97−48=49.
The problems on this page are the property of the MAA's American Mathematics Competitions