2012 AMC 10B Problems/Problem 20

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 adds $50$ 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

Let's test each number starting from $1$ , for our first case $1$ , we start out with $1$ and the number is then given to Bernardo. He will double the given number so in this case $1\times 2=2$ . So now he gives this resulting number to Silvia and she adds $50$ to the number. So we have $2+50=52$ . Now this is then passed to Bernardo so he doubles this $52$ and we have $52\times 2=104$ . And now this is then given to Silvia so $104+50=154$ . So she passes it to Bernardo and we have $154\times 2=308$ . Silvia gets it and so we have $308+50=358$ . Continuing this we have $358\times 2=716$ . There is no need to continue with this because when Silvia adds $50$

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2012 AMC 10B Problems/Problem 20

Problem

Solution