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 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
Solution 1
The last number that Bernado says has to be between 950 and 999. Note that 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 . Silvia 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 .
Solution 3
If our first number is then the sequence of numbers will be Note that we cannot go any further because doubling gives an extra at the end, which is already greater than The smallest will be given if Since the problem asks for the smallest possible value of we get and the sum of its digits is
~ 1001
See Also
2012 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 10 Problems and Solutions |
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 |
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