2012 AMC 12B Problems/Problem 14
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 addes 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
Solution 1
The last number that Bernado says has to be between 950 and 999. Note that 1->2->52->104->154->308->358->716->776 contains 4 doubling actions. Thus, we have .
Thus, . Then, . If , we have . Working backwards from 956,
.
So the starting number is 16, and our answer is , which is A.
Solution 2
Work backwards. The last number Bernardo produces must be in the range . That means that before this, Silvia must produce a number in the range . Before this, Bernardo must produce a number in the range . Before this, Silvia must produce a number in the range . Before this, Bernardo must produce a number in the range . Before this, Silvia must produce a number in the range . Before this, Bernardo must produce a number in the range . Before this, Silvia must produce a number in the range . Bernardo could not have added 50 to any number before this to obtain a number in the range , hence the minimum is 16 with the sum of digits being .
See Also
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |