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2007 AMC 10B Problem 9

2007 AMC 10B Problem 9AMC 10 Step by Step

Name: Date: September 13, 2026

Modular Arithmetic · difficulty 2 of 5 · about 3 min · position P1-10

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 $\text{"Lee's sis is a Mississippi miss, Chriss!"?}$

A)
$g$
B)
$h$
C)
$o$
D)
$s$
E)
$t$

Answer

Key insight

The final s is the 12th s in the message, so it shifts by 1 + 2 + ... + 12 = 78, a multiple of 26, and stays s.

Solution

First count how many times the letter $s$ appears, word by word:

word	Lee's	sis	is	a	Mississippi	miss	Chriss
number of $s$	1	2	1	0	4	2	2

Running total: $1, 3, 4, 4, 8, 10, 12$ . The final $s$ is the $12$ th occurrence.

The $k$ th occurrence of a letter moves $1 + 2 + \cdots + k = \frac{k(k+1)}{2}$ places to the right. For $k = 12$ this is $\frac{12 \cdot 13}{2} = 78$ .

Since the alphabet has $26$ letters, a shift of $78 = 3 \cdot 26$ brings the letter all the way around three full times and back to where it started. The last $s$ is replaced by $s$ itself.

The answer is $\boxed{\textbf{(D)}\ s}$ .

Why this works

A cyclic alphabet means shifts only matter modulo $26$ , so the question reduces to a triangular number mod $26$ . The real risk is clerical: count the occurrences carefully, and reduce the shift before trying to step through the alphabet.

The trap

Miscounting the number of s's (the two in 'Mississippi' pairs and the double s in 'Chriss' are easy to drop) and shifting by the wrong triangular number.

Common mistakes

Miscounting the number of s's (the two in 'Mississippi' pairs and the double s in 'Chriss' are easy to drop) and shifting by the wrong triangular number.
Shifting by $12$ rather than by $1 + 2 + \cdots + 12$ , which lands on $e$ and is not even a choice.

Techniques

Set up the equation/formula and compute; no special trick needed