HMMT 二月 2010 · 冲刺赛 · 第 6 题
HMMT February 2010 — Guts Round — Problem 6
题目详情
- [ 5 ] How many different numbers are obtainable from five 5 s by first concatenating some of the 5 s, thenmultiplying them together? For example, we could do 5 · 55 · 55, 555 · 55, or 55555, but not 5 · 5 or 2525.
13 ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUNDth
英文原题
[ 5 ] How many different numbers are obtainable from five 5s by first concatenating some of the 5s, then
multiplying them together? For example, we could do 5 · 55 · 55, 555 · 55, or 55555, but not 5 · 5 or 2525.
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13 th ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND
解析
英文解析
- [ 5 ] How many different numbers are obtainable from five 5 s by first concatenating some of the 5 s, thenmultiplying them together? For example, we could do 5 · 55 · 55, 555 · 55, or 55555, but not 5 · 5 or 2525.
Answer: 7 If we do 55555, then we’re done.
Note that 5, 55, 555, and 5555 all have completely distinguishable prime factorizations. This meansthat if we are given a product of them, we can obtain the individual terms. The number of 5555’s isthe exponent of 101, the number of 555’s is the exponent of 37, the number of 55’s is the exponentof 11 minus the exponent of 101, and the number of 5’s is just whatever we need to get the properexponent of 5. Then the answer is the number of ways we can split the five 5’s into groups of at leastone. This is the number of unordered partitions of 5, which is 7.
13 ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUNDth