2012 AMC 12B Problem 14

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

Small hint:

Bernardo wins on a turn when his doubled result reaches

1000,

forcing Silvia’s next number to be at least

1000

Big hint:

Work backwards: find the smallest start that first reaches

\ge1000

after Bernardo’s move, over one, two, three, or four rounds

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.