HMMT 二月 2012 · TEAM2 赛 · 第 5 题

HMMT February 2012 — TEAM2 Round — Problem 5

[ 10 ] Steph and Jeff each start with the number 4, and Travis is flipping coins. Every time he flips a heads, Steph replaces her number x with 2 x − 1, and Jeff replaces his number y with y + 8. Every x +1 time he flips a tails, Steph replaces her number x with , and Jeff replaces his number y with y − 3. 2 After some (positive) number of coin flips, Steph and Jeff miraculously end up with the same number below 2012. How many times was a coin flipped? For problems 6-10, justification is required.

解析

[ 10 ] Steph and Jeff each start with the number 4, and Travis is flipping a coin. Every time he flips a heads, Steph replaces her number x with 2 x − 1, and Jeff replaces his number y with y + 8. Every x +1 time he flips a tails, Steph replaces her number x with , and Jeff replaces his number y with y − 3. 2 After some (positive) number of coin flips, Steph and Jeff miraculously end up with the same number below 2012. How many times was the coin flipped? Answer: 137 Suppose that a heads and b tails are flipped. Jeff’s number at the end is 4 + 8 a − 3 b . Note that the operations which Steph applies are inverses of each other, and as a result it is not difficult a − b to check by induction that her final number is simply 1 + 3 · 2 . a − b n We now have 3 + 3( a − b ) + 5 a = 3 · 2 . Letting n = a − b , we see that 2 − n − 1 must be divisible by n 5, so that a is an integer. In particular, n is a positive integer. Furthermore, we have 1 + 3 · 2 < 2012, so that n ≤ 9. We see that the only possibility is for n = 7 = a − b , and thus 4 + 8 a − 3 b = 385. Solving, we get a = 72 , b = 65, so our answer is 72 + 65 = 137.