Difference between revisions of "2012 AMC 12B Problems/Problem 14"

Revision as of 12:47, 20 January 2014

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 addes $50$ 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 Bernado 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, $16x>250 \implies x \geq 16$ . If $x=16$ , we have $16x+700=956$ . Working backwards from 956,

$956 \rightarrow 478 \rightarrow 428 \rightarrow 214 \rightarrow 164 \rightarrow 82 \rightarrow 32 \rightarrow 16$ .

So the starting number is 16, and our answer is $1+6=\boxed{7}$ , which is A.

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 $[16,18]$ . Bernardo 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 $\boxed{7}$ .

2012 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 8: / Line 8: @@
 === Solution 1 ===
-The last number that Bernado says has to be between 950 and 999. Note that <math>1\rightarrow 2\rightarrow 52\rightarrow 104\rightarrow 154\rightarrow 308\rightarrow 358\rightarrow 716\rightarrow 776</math> contains 4 doubling actions. Thus, we have <math>x \rightarrow 2x \rightarrow 2x+50 \rightarrow 4x+100 \rightarrow 4x+150 \rightarrow 8x+300 \rightarrow 8x+350 \rightarrow 16x+700</math>.
+The last number that Bernado says has to be between 950 and 999. Note that <math>1\rightarrow 2\rightarrow 52\rightarrow 104\rightarrow 154\rightarrow 308\rightarrow 358\rightarrow 716\rightarrow 766</math> contains 4 doubling actions. Thus, we have <math>x \rightarrow 2x \rightarrow 2x+50 \rightarrow 4x+100 \rightarrow 4x+150 \rightarrow 8x+300 \rightarrow 8x+350 \rightarrow 16x+700</math>.
 Thus, <math>950<16x+700<1000</math>. Then, <math>16x>250 \implies x \geq 16</math>. If <math>x=16</math>, we have <math>16x+700=956</math>. Working backwards from 956,

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