Maths Olympiad Prep

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Algebra Difficulty 5.7 AIME, harder Prove it United States

Problem:

Steph and Jeff each start with the number 44, and Travis is flipping a coin. Every time he flips a heads, Steph replaces her number xx with 2x12x-1, and Jeff replaces his number yy with y+8y+8. Every time he flips a tails, Steph replaces her number xx with x+12\frac{x+1}{2}, and Jeff replaces his number yy with y3y-3. After some (positive) number of coin flips, Steph and Jeff miraculously end up with the same number below 20122012. How many times was the coin flipped?

Solution

Solution:

Answer: 137137

Suppose that aa heads and bb tails are flipped. Jeff's number at the end is 4+8a3b4 + 8a - 3b. Note that the operations which Steph applies are inverses of each other, and as a result it is not difficult to check by induction that her final number is simply 1+32ab1 + 3 \cdot 2^{a-b}.

We now have 4+8a3b=1+32ab4 + 8a - 3b = 1 + 3 \cdot 2^{a-b}. Letting n=abn = a-b, we see that 2nn12^n - n - 1 must be divisible by 55, so that aa is an integer. In particular, nn is a positive integer. Furthermore, we have 1+32n<20121 + 3 \cdot 2^n < 2012, so that n9n \leq 9. We see that the only possibility is for n=7=abn = 7 = a-b, and thus 4+8a3b=3854 + 8a - 3b = 385. Solving, we get a=72a = 72, b=65b = 65, so our answer is 72+65=13772 + 65 = 137.

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