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HMMT 二月 2017 · ALGNT 赛 · 第 8 题

HMMT February 2017 — ALGNT Round — Problem 8

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. Consider all ordered pairs of integers ( a, b ) such that 1 ≤ a ≤ b ≤ 100 and
    ( a + b )( a + b + 1)
    is an integer.ab
    Among these pairs, find the one with largest value of b . If multiple pairs have this maximal value ofb , choose the one with largest a . For example choose (3 , 85) over (2 , 85) over (4 , 84). Note that youranswer should be an ordered pair.
解析

英文解析

  1. Consider all ordered pairs of integers ( a, b ) such that 1 ≤ a ≤ b ≤ 100 and
    ( a + b )( a + b + 1)
    is an integer.ab
    Among these pairs, find the one with largest value of b . If multiple pairs have this maximal value ofb , choose the one with largest a . For example choose (3 , 85) over (2 , 85) over (4 , 84). Note that youranswer should be an ordered pair.
    Proposed by: Alexander Katz
    Answer: (35,90)
    2 2
    ( a + b )( a + b +1)
    a + b + a + b
    Firstly note that = 2+ . Let c be this fraction so that ( a + b )( a + b +1) = ab ( c +2)
    ab abfor some integers a, b, c . Suppose ( a, b ) with a ≥ b is a solution for some c . Consider the quadratic
    2 2
    x − ( bc − 1) x + b + b = 0
    It has one root a , and the other root is therefore bc − a − 1. Furthermore the other root can also be
    2 2
    b + b b + bexpressed as ≤ = b , so that 0 < bc − a − 1 ≤ b . In particular, ( b, bc − a − 1) is a solution asa b +1
    well.
    Thus all solutions ( a, b ) reduce to a solution where a = b , at which point c = 2 + . Since a, c are 2
    positive integers we thus have a = 1 , 2, and so c = 3 , 4 .a
    Through this jumping process, we iteratively find the solutions for c = 3:
    (2 , 2) → (2 , 3) → (3 , 6) → (6 , 14) → (14 , 35) → (35 , 90)
    and c = 4:
    (1 , 2) → (2 , 6) → (6 , 21) → (21 , 77)
    so that the desired pair is (35 , 90) .