PUMaC 2010 · 数论（B 组） · 第 7 题

PUMaC 2010 — Number Theory (Division B) — Problem 7

Find the numerator of 2011 ones ︷︸︸︷ 1010 11 . . . 11 0101 1100 11 . . . 11 0011 ︸︷︷︸ 2011 ones when reduced. 2010

解析

Find the numerator of 2011 ones ︷︸︸︷ 1010 11 . . . 11 0101 1100 11 . . . 11 0011 ︸︷︷︸ 2011 ones when reduced. 2 4 5 2 n +1 2 n +2 2 n +4 2 n +6 2 3 Solution: Note that 1 + x + x + x + . . . + x + x + x + x = (1 − x + x − x + 4 2 n +1 2 n +2 4 5 2 n +1 2 n +2 2 n +5 2 n +6 x )(1+ x + . . . + x + x ), as well as 1+ x + x + x + . . . + x + x + x + x = 2 4 2 n +1 2 n +2 (1 − x + x )(1 + x + . . . + x + x ). The easiest way to see this is to either multiply 1 numerator and denominator by x − 1, or to just numerically plug in small odd values of 2011 9091 in the original equation. Plugging in n = 1006 and x = 10 gives the fraction as . 9901 2010