HMMT 二月 2009 · 冲刺赛 · 第 3 题
HMMT February 2009 — Guts Round — Problem 3
题目详情
- [ 5 ] Find all pairs of integer solutions ( n, m ) ton m
3 2
2 = 3 − 1 .
12 HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUNDth
0 1 2009
英文原题
[ 5 ] Find all pairs of integer solutions ( n, m ) to
2 3 n
= 3 2 m
− 1 .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12 th HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUND
解析
英文解析
- [ 5 ] Find all pairs of integer solutions ( n, m ) ton m
3 2
2 = 3 − 1 .
Answer: (0 , 0) and (1 , 1)
x y
Solution: We find all solutions of 2 = 3 − 1 for positive integers x and y . If x = 1, we obtain the solution x = 1 , y = 1, which corresponds to ( n, m ) = (0 , 0) in the original problem. If x > 1, considerthe equation modulo 4. The left hand side is 0, and the right hand side is ( − 1) − 1, so y is even. Thusyx z z zwe can write y = 2 z for some positive integer z , and so 2 = (3 − 1)(3 + 1). Thus each of 3 − 1 andz
3 + 1 is a power of 2, but they differ by 2, so they must equal 2 and 4 respectively. Therefore, theonly other solution is x = 3 and y = 2, which corresponds to ( n, m ) = (1 , 1) in the original problem.
12 HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUNDth
0 1 2009