2006 AIME II Problem 3

Below is the professionally curated solution for Problem 3 of the 2006 AIME II, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AIME II solutions, or check the answer key.

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Concepts:prime factorizationdivisibility

Difficulty rating: 2150

3.

Let PP be the product of the first 100100 positive odd integers. Find the largest integer kk such that PP is divisible by 3k.3^k.

Solution:

P=135199,P = 1 \cdot 3 \cdot 5 \cdots 199, so kk is the total number of factors of 33 among the odd numbers up to 199.199. The odd multiples of 33 are 31,33,,365,3 \cdot 1, 3 \cdot 3, \ldots, 3 \cdot 65, and there are 3333 of them. The odd multiples of 99 are 91,,921:9 \cdot 1, \ldots, 9 \cdot 21: 1111 of them. The odd multiples of 2727 are 27,81,135,189:27, 81, 135, 189: 44 of them. The only odd multiple of 8181 at most 199199 is 8181 itself, and there are no multiples of 243.243.

Each layer contributes one additional factor of 3,3, so k=33+11+4+1=49.k = 33 + 11 + 4 + 1 = 49.

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