Difference between revisions of "2012 AMC 12B Problems/Problem 14"
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=== Solution 1 === | === Solution 1 === | ||
− | The last number that Bernado says has to be between 950 and 999. Note that 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>. |
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, | 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, |
Revision as of 15:55, 15 January 2013
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 addes
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 Bernado says has to be between 950 and 999. Note that contains 4 doubling actions. Thus, we have
.
Thus, . Then,
. If
, we have
. Working backwards from 956,
.
So the starting number is 16, and our answer is , which is A.
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
. Bernardo 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
.
See Also
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 |