Difference between revisions of "2012 AMC 10B Problems/Problem 20"

(Redirected page to 2012 AMC 12B Problems/Problem 14)
 
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== Problem ==
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#REDIRECT [[2012 AMC 12B Problems/Problem 14]]
 
 
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>?
 
 
 
<math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11 </math>
 
 
 
==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>
 

Latest revision as of 21:31, 28 February 2012