HMMT 二月 2017 · ALGNT 赛 · 第 8 题
HMMT February 2017 — ALGNT Round — Problem 8
题目详情
英文原题
- 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
- 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) .