Writing digits to reach a multiple of nine

Question

Two players build a six-digit number by writing one digit at a time, left to right, alternating, with the first player writing the first digit (which must not be 0). The first player wants the finished number to be divisible by 9; the second player wants it not to be. 1. Who wins with best play? 2. Now swap the goals: the first player wants the number not divisible by 9 and the second player wants it divisible. Who wins?

Accepted Answer

In both versions, the second player, who writes the last digit, wins. The divisibility test. A number is divisible by 9 if and only if the sum of its digits is divisible by 9. So the whole game is about the digit sum, and the last digit written decides it. Version 1: second player wants "not divisible". When the sixth digit is about to be written, the sum of the first five digits is some number s. The second player must choose a digit d from 0 to 9 so that s + d is not a multiple of 9. Among the ten possible digits, at most two make s + d a multiple of 9 (two digits that differ by 9, namely 0…

Writing digits to reach a multiple of nine

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