Difference between revisions of "2012 AMC 12B Problems/Problem 14"
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== Problem== | == Problem== | ||
− | Bernardo and Silvia play the following game. An integer between <math>0</math> and <math>999</math> 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 | + | Bernardo and Silvia play the following game. An integer between <math>0</math> and <math>999</math> 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 <math>50</math> to it and passes the result to Bernardo. The winner is the last person who produces a number less than <math>1000</math>. Let <math>N</math> be the smallest initial number that results in a win for Bernardo. What is the sum of the digits of <math>N</math>? |
<math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11 </math> | <math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11 </math> | ||
− | == Solution== | + | ==Solution== |
=== Solution 1 === | === Solution 1 === | ||
− | The last number that | + | 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>. |
− | Thus, <math>950<16x+700<1000</math>. Then, <math>16x>250 \implies x \geq 16</math>. | + | Thus, <math>950<16x+700<1000</math>. Then, <math>16x>250 \implies x \geq 16</math>. |
− | <math> | + | 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. |
− | + | ~minor edits by [https://artofproblemsolving.com/wiki/index.php/User:Kevinchen_yay KevinChen_Yay] | |
+ | |||
+ | ~removed unnecessary last step (EvanDu168) | ||
=== Solution 2 === | === Solution 2 === | ||
− | Work backwards. The last number Bernardo produces must be in the range <math>[950,999]</math>. That means that before this, Silvia must produce a number in the range <math>[475,499]</math>. Before this, Bernardo must produce a number in the range <math>[425,449]</math>. Before this, Silvia must produce a number in the range <math>[213,224]</math>. Before this, Bernardo must produce a number in the range <math>[163,174]</math>. Before this, Silvia must produce a number in the range <math>[82,87]</math>. Before this, Bernardo must produce a number in the range <math>[32,37]</math>. Before this, Silvia must produce a number in the range <math>[16,18]</math>. | + | Work backwards. The last number Bernardo produces must be in the range <math>[950,999]</math>. That means that before this, Silvia must produce a number in the range <math>[475,499]</math>. Before this, Bernardo must produce a number in the range <math>[425,449]</math>. Before this, Silvia must produce a number in the range <math>[213,224]</math>. Before this, Bernardo must produce a number in the range <math>[163,174]</math>. Before this, Silvia must produce a number in the range <math>[82,87]</math>. Before this, Bernardo must produce a number in the range <math>[32,37]</math>. Before this, Silvia must produce a number in the range <math>[16,18]</math>. Silvia could not have added 50 to any number before this to obtain a number in the range <math>[16,18]</math>, hence the minimum <math>N</math> is 16 with the sum of digits being <math>\boxed{\textbf{(A)}\ 7}</math>. |
+ | |||
+ | === Solution 3 === | ||
+ | |||
+ | If our first number is <math>N,</math> then the sequence of numbers will be <cmath>2N,~2N+50,~4N+100,~4N+150,~8N+300,~8N+350,~16N+700,~16N+750</cmath> Note that we cannot go any further because doubling gives an extra <math>1500</math> at the end, which is already greater than <math>1000.</math> The smallest <math>N</math> will be given if <math>16N+750>1000>16N+700 \implies 15<N<19.</math> Since the problem asks for the smallest possible value of <math>N,</math> we get <math>16,</math> and the sum of its digits is <math>1+6=\boxed{\textbf{(A)}~7}</math> | ||
+ | |||
+ | ~ 1001 | ||
== See Also == | == See Also == | ||
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{{AMC10 box|year=2012|ab=B|num-b=19|num-a=21}} | {{AMC10 box|year=2012|ab=B|num-b=19|num-a=21}} | ||
{{AMC12 box|year=2012|ab=B|num-b=13|num-a=15}} | {{AMC12 box|year=2012|ab=B|num-b=13|num-a=15}} | ||
+ | {{MAA Notice}} |
Latest revision as of 09:46, 24 April 2024
Problem
Bernardo and Silvia play the following game. An integer between and
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
to it and passes the result to Bernardo. The winner is the last person who produces a number less than
. Let
be the smallest initial number that results in a win for Bernardo. What is the sum of the digits of
?
Solution
Solution 1
The last number that Bernardo says has to be between 950 and 999. Note that contains 4 doubling actions. Thus, we have
.
Thus, . Then,
.
Because we are looking for the smallest integer ,
. Our answer is
, 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 . That means that before this, Silvia must produce a number in the range
. Before this, Bernardo must produce a number in the range
. Before this, Silvia must produce a number in the range
. Before this, Bernardo must produce a number in the range
. Before this, Silvia must produce a number in the range
. Before this, Bernardo must produce a number in the range
. Before this, Silvia must produce a number in the range
. Silvia could not have added 50 to any number before this to obtain a number in the range
, hence the minimum
is 16 with the sum of digits being
.
Solution 3
If our first number is then the sequence of numbers will be
Note that we cannot go any further because doubling gives an extra
at the end, which is already greater than
The smallest
will be given if
Since the problem asks for the smallest possible value of
we get
and the sum of its digits is
~ 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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