2012 AMC 10B Problem 20

Problem 20 of 25HarderAlgebra Number Theory

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?

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Solution

If the initial number is

x,

Bernardo’s successive outputs are

\begin{gathered} 2x,\quad4x+100,\quad8x+300,\\ 16x+700,\quad32x+1500. \end{gathered}

Silvia’s output after each of these is

50

larger. Bernardo wins on a given turn exactly when his output is below

1000

but the following output from Silvia is at least

1000.

For the first four Bernardo turns, the smallest integer

x

satisfying those two inequalities is, respectively,

475,213,82,

and

16.

A fifth Bernardo output is already at least

1500.

Thus the smallest winning initial number is

N=16,

whose digit sum is

1+6=7.

Thus, the correct answer is A .

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