2006 AIME II Problems/Problem 3
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 2
We count the multiples of below 200 and subtract the count of multiples of :
Solution 3
We can use a modified version of Legendre's Formula. First, we count the number of multiples of 3 in the sequence 1, 3, 5, 7, 9, ..., 195, 197, 199.
This is the same as the number of multiples of 3 in the sequence 3, 9, 15, 21, ..., 192, 195. There are clearly 33 terms in this sequence.
Next, we count the number of multiples of 9 in the sequence 1, 3, 5, 7, 9, ..., 195, 197, 199. This is the same as the number of multiples of 9 in the sequence 9, 18, 27, 36, ...., 189, 198 - but there's a catch. Note that every other member of this sequence isn't odd and thus is not part of the product of the first 100 odd integers, so our new sequence is actually 9, 27, 45...189. Divide every term by 9 to get a new sequence; 1, 3, 5...21, which clearly has 11 terms.
Next, we similarly count the number of multiples of 27 in the sequence 1, 3, 5, 7, 9, ..., 195, 197, 199. This is just 27, 81, 135, 189, so 4 multiples here.
Finally, we count the number of multiples of 81 in the sequence 1, 3, 5, 7, 9, ..., 195, 197, 199. There is only one such multiple, 81.
Any power of 3 above 81 doesn't fit into our sequence.
Finally, we have 33+11+4+1=49.
Our final answer is 49.
(Note that I oversimplified this a lot, in real life we wouldn't have to list out the sequences as tediously as I did).
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 | ||
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