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2012 AMC 12B Problem 14

Problem 14 of 25IntermediateAlgebraProblem-Solving Techniques

Bernardo and Silvia play the following game. An integer between 00 and 999,999, inclusive, is selected and given to Bernardo. Whenever Bernardo receives a number, he doubles it and passes the result to Silvia. Whenever Silvia receives a number, she adds 5050 to it and passes the result to Bernardo. The winner is the last person who produces a number less than 1000.1000. Let NN be the smallest initial number that results in a win for Bernardo. What is the sum of the digits of N?N?

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Solution

Bernardo wins after a round when his doubled number 2n+50≥10002n+50\ge1000 but the previous numbers stayed below 1000.1000. The smallest nn with 2n+50≥10002n+50\ge1000 is 475.475. Working backwards, the smallest starting values that lead to a win after two, three, and four rounds are the smallest integers with 2n+50≥475,2n+50\ge475, ≥213,\ge213, and ≥82,\ge82, namely 213,213, 82,82, and 16.16. No start wins after more than four rounds. So N=16,N=16, and the sum of its digits is 1+6=7.1+6=7. Thus, the correct answer is A.
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