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Difference between revisions of "2007 AMC 10B Problems/Problem 9"

(Created page with '==Problem 9== 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 <math>1</math> place to its ri…')
 
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==Problem 9==
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==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 <math>1</math> place to its right in the alphabet (asumming that the letter <math>A</math> is one place to the right of the letter <math>Z</math>). The second time this same letter appears in the given message, it is replaced by the letter that is <math>1+2</math> places to the right, the third time it is replaced by the letter that is <math>1+2+3</math> places to the right, and so on. For example, with this code the word "banana" becomes "cbodqg". What letter will replace the last letter <math>s</math> in the message
 
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 <math>1</math> place to its right in the alphabet (asumming that the letter <math>A</math> is one place to the right of the letter <math>Z</math>). The second time this same letter appears in the given message, it is replaced by the letter that is <math>1+2</math> places to the right, the third time it is replaced by the letter that is <math>1+2+3</math> places to the right, and so on. For example, with this code the word "banana" becomes "cbodqg". What letter will replace the last letter <math>s</math> in the message
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<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>
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== See Also ==
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{{AMC10 box|year=2007|ab=B|num-b=8|num-a=10}}

Revision as of 16:08, 4 June 2011

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 $1$ place to its right in the alphabet (asumming that the letter $A$ is one place to the right of the letter $Z$). The second time this same letter appears in the given message, it is replaced by the letter that is $1+2$ places to the right, the third time it is replaced by the letter that is $1+2+3$ places to the right, and so on. For example, with this code the word "banana" becomes "cbodqg". What letter will replace the last letter $s$ in the message \[\text{"Lee's sis is a Mississippi miss, Chriss!"?}\]

$\textbf{(A) } g \qquad\textbf{(B) } h \qquad\textbf{(C) } o \qquad\textbf{(D) } s \qquad\textbf{(E) } t$

Solution

Since the letter that will replace the last $s$ does not depend on any letter except the other $s$'s, you can disregard anything else. There are $12$ $s$'s, so the last $s$ will be replaced the letter $1+2+3+\cdots+12$ places to the right of $s$. \[1+2+3+\cdots+12=12 \times \frac{1+12}{2} = 78\] Every $26$ places, you will end up with the same letter so you can just take the remainder of $78$ when you divide by $26,$ which is $0.$ Therefore, the letter that will is replace the last $s$ is $\boxed{\textbf{(D) } s}$

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
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