Difference between revisions of "2007 AMC 10B Problems/Problem 9"
m (→Solution) |
|||
| Line 8: | Line 8: | ||
==Solution== | ==Solution== | ||
| − | Since the letter that will replace the last <math>s</math> does not depend on any letter except the other <math>s</math>'s, you can disregard anything else. There are <math>12</math> <math>s</math>'s, so the last <math>s</math> will be replaced the letter <math>1+2+3+\cdots+12</math> places to the right of <math>s</math>. | + | Since the letter that will replace the last <math>s</math> does not depend on any letter except the other <math>s</math>'s, you can disregard anything else. There are <math>12</math> <math>s</math>'s, so the last <math>s</math> will be replaced by the letter <math>1+2+3+\cdots+12</math> places to the right of <math>s</math>. |
<cmath>1+2+3+\cdots+12=12 \times \frac{1+12}{2} = 78</cmath> | <cmath>1+2+3+\cdots+12=12 \times \frac{1+12}{2} = 78</cmath> | ||
Every <math>26</math> places, you will end up with the same letter so you can just take the remainder of <math>78</math> when you divide by <math>26,</math> which is <math>0.</math> Therefore, the letter that will is replace the last <math>s</math> is <math>\boxed{\textbf{(D) } s}</math> | Every <math>26</math> places, you will end up with the same letter so you can just take the remainder of <math>78</math> when you divide by <math>26,</math> which is <math>0.</math> Therefore, the letter that will is replace the last <math>s</math> is <math>\boxed{\textbf{(D) } s}</math> | ||
Revision as of 12:05, 4 June 2021
Problem
A cryptographic code is designed as follows. The first time a letter appears in a given message it is replaced by the letter that is 
place to its right in the alphabet (asumming that the letter 
is one place to the right of the letter 
). The second time this same letter appears in the given message, it is replaced by the letter that is 
places to the right, the third time it is replaced by the letter that is 
places to the right, and so on. For example, with this code the word "banana" becomes "cbodqg". What letter will replace the last letter 
in the message
![\[\text{"Lee's sis is a Mississippi miss, Chriss!"?}\]](http://latex.artofproblemsolving.com/b/4/9/b490ec780d203abe9592b8a926abcede62c6bd10.png)
Solution
Since the letter that will replace the last does not depend on any letter except the other 's, you can disregard anything else. There are 
's, so the last will be replaced by the letter 
places to the right of .
![\[1+2+3+\cdots+12=12 \times \frac{1+12}{2} = 78\]](http://latex.artofproblemsolving.com/f/6/6/f666aa9cb31759468bf8dfb5efcb113f6cf17c1a.png)
Every 
places, you will end up with the same letter so you can just take the remainder of 
when you divide by 
which is 
Therefore, the letter that will is replace the last is 
See Also
| 2007 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
