HMMT 二月 2009 · 冲刺赛 · 第 33 题
HMMT February 2009 — Guts Round — Problem 33
题目详情
- [ 18 ] Let m be a positive integer. Let d ( n ) denote the number of divisors of n , and define the function m 105 ∑ d ( n ) F ( x ) = . x n n =1 Define the numbers a ( n ) to be the positive integers for which 2 m 105 ∑ a ( n ) 2 F ( x ) = x n n =1 m for all real x . Express a (105 ) in terms of m . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 12 HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUND
解析
- [ 18 ] Let m be a positive integer. Let d ( n ) denote the number of divisors of n , and define the function m 105 ∑ d ( n ) F ( x ) = . x n n =1 Define the numbers a ( n ) to be the positive integers for which 2 m 105 ∑ a ( n ) 2 F ( x ) = x n n =1 m for all real x . Express a (105 ) in terms of m . ( ) 3 ( ) 3 2 3 m +3 m +6 m +11 m +6 Answer: OR 6 3 1 2 3 4 5 6 7 (The expanded polynomial (216+1188 m +2826 m +3815 m +3222 m +1767 m +630 m +141 m + 216 8 9 18 m + m ) is also an acceptable answer.) 2 x Solution: The denominator of a term in the expansion of F ( x ) is equal to n if and only if it is d ( n/k ) d ( k ) m a product of two terms of F of the form and for some divisor k of n . Thus a (105 ) = x x ( n/k ) k ∑ m 105 a b c m d ( k ) d ( ). We can write k = 3 5 7 with a, b, c ≤ m for any divisor k of 105 , and in this m k | 105 k case d ( k ) = ( a + 1)( b + 1)( c + 1). Thus the sum becomes ∑ ( a + 1)( b + 1)( c + 1)( m − a + 1)( m − b + 1)( m − c + 1) . 0 ≤ a,b,c ≤ m For a fixed b and c , we can factor out ( b + 1)( c + 1)( m − b + 1)( m − c + 1) from the terms having this b and c and find that the sum is equal to ( ) m +1 ∑ ∑ m a (105 ) = ( b + 1)( c + 1)( m − b + 1)( m − c + 1) a ( m − a + 2) a =1 0 ≤ b,c ≤ m ( ) ∑ ( m + 1)( m + 2) ( m + 1)( m + 2)(2 m + 3) = ( b + 1)( c + 1)( m − b + 1)( m − c + 1) ( m + 2) − 2 6 0 ≤ b,c ≤ m ( ) ∑ (3 m + 6 − 2 m − 3)( m + 1)( m + 2) = ( b + 1)( c + 1)( m − b + 1)( m − c + 1) 6 0 ≤ b,c ≤ m ( ) ∑ m + 3 = ( b + 1)( c + 1)( m − b + 1)( m − c + 1) 3 0 ≤ b,c ≤ m 8 ( ) 3 m +3 m Fixing c and factoring out terms again, we find by a similar argument that a (105 ) = . 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 12 HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUND