2012 AMC 12B Problems/Problem 14

Problem

Bernardo and Silvia play the following game. An integer between 0 and 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 addes 50 to it and passes the result to Bernardo. The winner is the last person who produces a number less than 1000. Let N be the smallest initial number that results in a win for Bernardo. What is the sum of the digits of N $\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$

Solution

The last number that Bernado says has to be between 950 and 999. Note that 1->2->52->104->154->308->358->716->776 contains 4 doubling actions. Thus, we have $x \rightarrow 2x \rightarrow 2x+50 \rightarrow 4x+100 \rightarrow 4x+150 \rightarrow 8x+300 \rightarrow 8x+350 \rightarrow 16x+700$ .

Thus, $950<16x+700<1000$ . Then, $16x>250 \implies x \geq 16$ . If $x=16$ , we have $16x+700=956$ . Working backwards from 956,

$956 \rightarrow 478 \rightarrow 428 \rightarrow 214 \rightarrow 164 \rightarrow 82 \rightarrow 32 \rightarrow 16$ .

So the starting number is 16, and our answer is $1+6=\boxed{7}$ , which is A.

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2012 AMC 12B Problems/Problem 14

Problem

Solution