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Difference between revisions of "2012 AMC 10B Problems/Problem 20"
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==Solution==
 
==Solution==
  
−
Let's test each solution, for our first case <math>7</math>, we start out with <math>7</math> and the number is then given to Bernardo. He will double the given number so in this case <math>7\times 2=14</math>. So now he gives this resulting number to Silvia and she adds <math>50</math> to the number. So we have <math>14+50=64</math>
+
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>

Revision as of 21:19, 28 February 2012

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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