# 2012 AMC 12B Problems/Problem 14

## Problem

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 $N$?

$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$

## Solution

### Solution 1

The last number that Bernardo says has to be between 950 and 999. Note that $1\rightarrow 2\rightarrow 52\rightarrow 104\rightarrow 154\rightarrow 308\rightarrow 358\rightarrow 716\rightarrow 766$ contains 4 doubling actions. Thus, we have $x \rightarrow 2x \rightarrow 2x+50 \rightarrow 4x+100 \rightarrow 4x+150 \rightarrow 8x+300 \rightarrow 8x+350 \rightarrow 16x+700$.

Thus, $950<16x+700<1000$. Then, $16x>250 \implies x \geq 16$.

Because we are looking for the smallest integer $x$, $x=16$. Our answer is $1+6=\boxed{7}$, which is A.

~minor edits by KevinChen_Yay

~removed unnecessary last step (EvanDu168)

### Solution 2

Work backwards. The last number Bernardo produces must be in the range $[950,999]$. That means that before this, Silvia must produce a number in the range $[475,499]$. Before this, Bernardo must produce a number in the range $[425,449]$. Before this, Silvia must produce a number in the range $[213,224]$. Before this, Bernardo must produce a number in the range $[163,174]$. Before this, Silvia must produce a number in the range $[82,87]$. Before this, Bernardo must produce a number in the range $[32,37]$. Before this, Silvia must produce a number in the range $[16,18]$. Silvia could not have added 50 to any number before this to obtain a number in the range $[16,18]$, hence the minimum $N$ is 16 with the sum of digits being $\boxed{\textbf{(A)}\ 7}$.

### Solution 3

If our first number is $N,$ then the sequence of numbers will be $$2N,~2N+50,~4N+100,~4N+150,~8N+300,~8N+350,~16N+700,~16N+750$$ Note that we cannot go any further because doubling gives an extra $1500$ at the end, which is already greater than $1000.$ The smallest $N$ will be given if $16N+750>1000>16N+700 \implies 15 Since the problem asks for the smallest possible value of $N,$ we get $16,$ and the sum of its digits is $1+6=\boxed{\textbf{(A)}~7}$

~ 1001