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HMMT 十一月 2014 · 冲刺赛 · 第 9 题

HMMT November 2014 — Guts Round — Problem 9

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

  1. [ 7 ] Let f be a function from the nonnegative integers to the positive reals such that f ( x + y ) = f ( x ) · f ( y )
    holds for all nonnegative integers x and y . If f (19) = 524288 k , find f (4) in terms of k .
    HMMT NOVEMBER 2014, 15 NOVEMBER 2014 — GUTS ROUND
    Organization Team Team ID#

英文原题

[ 7 ] Let f be a function from the nonnegative integers to the positive reals such that f ( x + y ) = f ( x ) · f ( y )
holds for all nonnegative integers x and y . If f (19) = 524288 k , find f (4) in terms of k .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
HMMT NOVEMBER 2014, 15 NOVEMBER 2014 — GUTS ROUND
Organization Team Team ID#

解析

英文解析

  1. [ 7 ] Let f be a function from the nonnegative integers to the positive reals such that f ( x + y ) = f ( x ) · f ( y )
    holds for all nonnegative integers x and y . If f (19) = 524288 k , find f (4) in terms of k .
    4 / 19 n
    Answer: 16 k The given condition implies f ( mn ) = f ( m ) , so
    19 4
    f (4) = f (4 · 19) = f (19 · 4) = f (19)
    4 / 19
    and it follows that f (4) = 16 k .