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HMMT 二月 2009 · 冲刺赛 · 第 33 题

HMMT February 2009 — Guts Round — Problem 33

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

  1. [ 18 ] Let m be a positive integer. Let d ( n ) denote the number of divisors of n , and define the functionm
    ∑105
    d ( n )
    F ( x ) = .
    n =1 nx
    Define the numbers a ( n ) to be the positive integers for which
    2 m
    ∑105
    a ( n )
    F ( x ) =2
    n =1 nxmfor all real x . Express a (105 ) in terms of m .
    12 HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUNDth

英文原题

[ 18 ] Let m be a positive integer. Let d ( n ) denote the number of divisors of n , and define the function
F ( x ) =
105 m
∑
n =1
d ( n )
n x .
Define the numbers a ( n ) to be the positive integers for which
F ( x ) 2 =
105 2 m
∑
n =1
a ( n )
n x
for all real x . Express a (105 m ) in terms of m .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12 th HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUND

解析

英文解析

  1. [ 18 ] Let m be a positive integer. Let d ( n ) denote the number of divisors of n , and define the functionm
    ∑105
    d ( n )
    F ( x ) = .
    n =1 nx
    Define the numbers a ( n ) to be the positive integers for which
    2 m
    ∑105
    a ( n )
    F ( x ) =2
    n =1 nxmfor 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 +
    8 9216
    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 isd ( n/k ) d ( k )
    a product of two terms of F of the form and for some divisor k of n . Thus a (105 ) =mx x
    ( n/k ) k
    ∑
    105 ma b c md ( k ) d ( ). We can write k = 3 5 7 with a, b, c ≤ m for any divisor k of 105 , and in thismk | 105
    case d ( k ) = ( a + 1)( b + 1)( c + 1). Thus the sum becomesk
    ∑
    ( 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 thisb and c and find that the sum is equal to
    ( )
    m +1
    ∑ ∑
    a (105 ) = ( b + 1)( c + 1)( m − b + 1)( m − c + 1) a ( m − a + 2)ma =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)
    0 ≤ b,c ≤ m 6
    ( )
    ∑
    m + 3 = ( b + 1)( c + 1)( m − b + 1)( m − c + 1)
    0 ≤ b,c ≤ m 3
    ( )8
    m +33
    Fixing c and factoring out terms again, we find by a similar argument that a (105 ) = .m
    12 HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUNDth