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> | + | 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> |
Revision as of 21:11, 28 February 2012
Problem
Bernardo and Silvia play the following game. An integer between 
and 
, 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 
to it and passes the result to Bernardo. The winner is the last person who produces a number less than 
. Let 
be the smallest initial number that results in a win for Bernardo. What is the sum of the digits of ?
Solution
Let's test each solution, for our first case 
, we start out with and the number is then given to Bernardo. He will double the given number so in this case 
. So now he gives this resulting number to Silvia and she adds to the number. So we have 
