HMMT 十一月 2011 · 冲刺赛 · 第 15 题
HMMT November 2011 — Guts Round — Problem 15
题目详情
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4 ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 12 NOVEMBER 2011— GUTS ROUNDTH
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英文原题
[8]
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4 TH ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 12 NOVEMBER 2011— GUTS ROUND
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解析
英文解析
- [ 8 ] For positive integers n , let L ( n ) be the largest factor of n other than n itself. Determine the numberof ordered pairs of composite positive integers ( m, n ) for which L ( m ) L ( n ) = 80.
Answer: 12 Let x be an integer, and let p be the smallest prime factor of x . Then, if L ( a ) = x ,
we note that we must have a = px for some prime p ≤ p . (Otherwise, if p > p , then > x . If p ispxxx xcomposite, then kx > x for some factor k of x .)xp
So we have:
• L ( a ) = 2 , 4 , 8 , 10 , 16 , 20 , 40 ⇒ 1 value for a
• L ( a ) = 5 ⇒ 3 values for a
Hence, we note that, since m and n are composite, we cannot have L ( m ) = 1 or L ( n ) = 1, so thepossible pairs ( L ( m ) , L ( n )) are (2 , 40) , (4 , 20) , (5 , 16) , (8 , 10) and vice-versa.
We add the number of choices for each pair, and double since m and n are interchangeable, to get
2(1 × 1 + 1 × 1 + 3 × 1 + 1 × 1) = 12 possible ordered pairs ( m, n ).