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| − | == Problem ==
| + | #REDIRECT [[2012 AMC 12B Problems/Problem 14]] |
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| − | Bernardo and Silvia play the following game. An integer between <math>0</math> and <math>999</math>, 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 <math>50</math> to it and passes the result to Bernardo. The winner is the last person who produces a number less than <math>1000</math>. Let <math>N</math> be the smallest initial number that results in a win for Bernardo. What is the sum of the digits of <math>N</math>?
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| − | <math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11 </math>
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| − | ==Solution==
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| − | Let's test each number starting from <math>1</math>, for our first case <math>1</math>, we start out with <math>1</math> and the number is then given to Bernardo. He will double the given number so in this case <math>1\times 2=2</math>. So now he gives this resulting number to Silvia and she adds <math>50</math> to the number. So we have <math>2+50=52</math>. Now this is then passed to Bernardo so he doubles this <math>52</math> and we have <math>52\times 2=104</math>. And now this is then given to Silvia so <math>104+50=154</math>. So she passes it to Bernardo and we have <math>154\times 2=308</math>. Silvia gets it and so we have <math>308+50=358</math>. Continuing this we have <math>358\times 2=716</math>. There is no need to continue with this because when Silvia adds <math>50</math>
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