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 $N?$

$7$

$8$

$9$

$10$

$11$

Answer: A

Concepts:work backwards inequality

Difficulty rating: 1730

Solution:

Bernardo wins after a round when his doubled number $2n+50\ge1000$ but the previous numbers stayed below $1000.$ The smallest $n$ with $2n+50\ge1000$ is $475.$

Working backwards, the smallest starting values that lead to a win after two, three, and four rounds are the smallest integers with $2n+50\ge475,$ $\ge213,$ and $\ge82,$ namely $213,$ $82,$ and $16.$ No start wins after more than four rounds.

So $N=16,$ and the sum of its digits is $1+6=7.$

Thus, the correct answer is A.

2012 AMC 12B Problem 14

Answer: A

Difficulty rating: 1730

Solution:

Problem 14 in Other Years