Olympiad Maths Prep

Track / Stage 3 / 82 of 260 #82 of 2000

Problem 82

AMC 10/12, early questions
Combinatorics Difficulty 3.2 Find the answer

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 11 place to its right in the alphabet (asumming that the letter AA is one place to the right of the letter ZZ). The second time this same letter appears in the given message, it is replaced by the letter that is 1+21+2 places to the right, the third time it is replaced by the letter that is 1+2+31+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 ss in the message
"Lee’s sis is a Mississippi miss, Chriss!"?\text{"Lee's sis is a Mississippi miss, Chriss!"?}
(A) g(B) h(C) o(D) s(E) t\textbf{(A) } g \qquad\textbf{(B) } h \qquad\textbf{(C) } o \qquad\textbf{(D) } s \qquad\textbf{(E) } t

Official solution

Since the letter that will replace the last ss does not depend on any letter except the other ss's, you can disregard anything else. There are 1212 ss's, so the last ss will be replaced by the letter 1+2+3++121+2+3+\cdots+12 places to the right of ss.
1+2+3++12=12×1+122=781+2+3+\cdots+12=12 \times \frac{1+12}{2} = 78
Every 2626 places, you will end up with the same letter so you can just take the remainder of 7878 when you divide by 26,26, which is 0.0. Therefore, the letter that will replace the last ss is (D) s\boxed{\textbf{(D) } s}

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.