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

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The last number that Bernardo 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>.  
 
The last number that Bernardo 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>.  
  
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Thus, <math>950<16x+700<1000</math>. Then, <math>16x>250 \implies x \geq 16</math>.
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Thus, <math>950<16x+700<1000</math>. Then, <math>16x>250 \implies x \geq 16</math>.  
  
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So the starting number is 16, and our answer is <math>1+6=\boxed{7}</math>, which is A.
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Because we are looking for the smallest integer <math>x</math>, <math>x=16</math>. Our answer is <math>1+6=\boxed{7}</math>, which is A.
  
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~minor edits by [https://artofproblemsolving.com/wiki/index.php/User:Kevinchen_yay KevinChen_Yay]
  
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~minor edits by KevinChen_Yay
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~removed unnecessary last step (EvanDu168)
  
 
=== Solution 2 ===
 
=== Solution 2 ===

Latest revision as of 08:46, 24 April 2024

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 $16N+750>1000>16N+700 \implies 15<N<19.$ 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

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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