Question

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 $N?$

$7$

$8$

$9$

$10$

$11$

Solution:

Suppose $x$ was our initial number. Then, it becomes $2x$ when given to Silvia, and $2x+50$ when given to Bernardo. Repeatedly doing this can yield that it eventually becomes $16x+700$ when given to Silvia and $16x+750$ when given to Bernardo. Any more iterations makes the number greater than $1000.$

The number given to Silvia must be below $1000$ and the number Silvia makes is greater than $1000,$ so $16x+750 > 1000 > 16x+700 .$

Therefore, $300 > 16x > 250 .$ This makes $18 \geq x \geq 16,$ so $N=16.$ As such, the sum of the digits of $N$ is $1+6=7.$

Thus, the correct answer is A .

2012 AMC 10B Problem 20

Solution:

Problem 20 in Other Years