AMC 10 · 2007 · #9
Easy mode Grade 4A secret code slides each letter to the right in the alphabet. The alphabet loops around, so the letter after Z is A. The first time a letter shows up, it slides 1 place to the right. The second time that same letter shows up, it slides 1+2 places. The third time, 1+2+3 places, and so on. For example, the word "banana" turns into "cbodqg". In the message
"Lee’s sis is a Mississippi miss, Chriss!"
what letter takes the place of the very last s?
Pick an answer.
AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A letter gets shifted right in the alphabet by a growing amount each time it reappears: 1 place the first time, 1+2 places the second time, 1+2+3 places the third time, and so on, wrapping from Z back to A. In the message "Lee's sis is a Mississippi miss, Chriss!", find which letter replaces the very last s.
Givens: The n-th time a letter appears, it is replaced by the letter 1+2+...+n places to its right.; The alphabet is a 26-letter cycle: A sits one place to the right of Z.; The message is "Lee's sis is a Mississippi miss, Chriss!"; Example: "banana" becomes "cbodqg" (2nd n shifts 1+2=3 to q, 3rd a shifts 1+2+3=6 to g).
Unknowns: The letter that replaces the final s in the message
Understand
Restated: A letter gets shifted right in the alphabet by a growing amount each time it reappears: 1 place the first time, 1+2 places the second time, 1+2+3 places the third time, and so on, wrapping from Z back to A. In the message "Lee's sis is a Mississippi miss, Chriss!", find which letter replaces the very last s.
Givens: The n-th time a letter appears, it is replaced by the letter 1+2+...+n places to its right.; The alphabet is a 26-letter cycle: A sits one place to the right of Z.; The message is "Lee's sis is a Mississippi miss, Chriss!"; Example: "banana" becomes "cbodqg" (2nd n shifts 1+2=3 to q, 3rd a shifts 1+2+3=6 to g).
Plan
Primary tool: #5 Look for a Pattern
Secondary: #16 Change Focus / Count the Complement, #2 Make a Systematic List
The rule builds a clean pattern: the n-th s is pushed 1+2+...+n places right, a triangular number. And because the alphabet loops every 26 letters, only the shift's remainder after dividing by 26 matters. So the plan is to count the s's, find the running shift for the last one from the triangular pattern, then wrap it around the 26-letter cycle.
Execute — Answer: D
4.OA.A.3 Step 1 Only the s's matter
- The shift on any letter is set entirely by how many times that same letter has appeared before it.
- The last s's replacement therefore depends only on the other s's, not on the l, i, m, p, or any other letter.
- So strip the message down to just its s's and ignore everything else.
💡 Each letter runs its own private counter, so the s's never care about the other letters.
2.OA.B.2 Step 2 Count the s's
- Go through the message and tally every s (capital or small letters are the same letter): Lee's has 1, sis has 2 (total 3), is has 1 (total 4), Mississippi has 4 (total 8), miss has 2 (total 10), Chriss has 2 (total 12).
- So there are 12 s's, and the one we want is the 12th.
💡 Sweep left to right keeping a single running count so no s is missed or double-counted.
4.OA.C.5 Step 3 Find the 12th s's shift
- The 12th time a letter appears, it moves 1+2+3+...+12 places to the right.
- This running total is a triangular number, which equals the number of terms times one more than that, divided by 2.
- For 12 terms that is 12 times 13 divided by 2, which is 78.
- So the last s is shifted 78 places to the right.
💡 Pairing the first and last terms (1+12, 2+11, ...) makes the growing sum easy to total.
4.NBT.B.6 Step 4 Wrap around the 26-letter alphabet
- Because the alphabet has 26 letters and loops back to A after Z, shifting by a full 26 places lands on the same letter you started from.
- So only the remainder of 78 divided by 26 counts.
- Since 78 is exactly 3 times 26, the remainder is 0, meaning the letter does not move at all.
- The 12th s is replaced by s itself, so the answer is (D).
💡 Every full lap of 26 letters brings you right back home, so only the leftover part changes the letter.
4.OA.A.3 The shift on any letter is set entirely by how many times that same letter has a 2.OA.B.2 Go through the message and tally every s (capital or small letters are the same 4.OA.C.5 The 12th time a letter appears, it moves 1+2+3+...+12 places to the right. This 4.NBT.B.6 Because the alphabet has 26 letters and loops back to A after Z, shifting by a f Review
Reasonableness: A shift of 0 means the letter stays put, and s is one of the offered choices, so the result fits. The check also passes the sanity test on the given example: the 3rd a in "banana" shifts 1+2+3=6 places, giving g, exactly as "cbodqg" shows, so the same triangular-number rule that lands the last s on 78 is being applied correctly.
Alternative: Instead of the formula, list the running shift for each s in order: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78. The 12th entry is 78, and subtracting 26 twice (78, 52, 26, 0) shows it reduces to 0, again leaving s unchanged.
CCSS standards used (min grade 4)
4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Interpreting the coding rule and realizing the last s's replacement depends only on the earlier s's.)2.OA.B.2Fluently add and subtract within 20 using mental strategies (Tallying the message to count that s appears 12 times in all.)4.OA.C.5Generate a number or shape pattern following a given rule (Recognizing the shift 1+2+...+n as a triangular number and computing it as 78 for the 12th appearance.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Taking 78 modulo 26 to wrap the shift around the alphabet and finding the remainder 0.)
⭐ Each letter counts only its own appearances, and since the alphabet loops every 26 steps, a shift that is a multiple of 26 leaves the letter unchanged.
⭐ Each letter counts only its own appearances, and since the alphabet loops every 26 steps, a shift that is a multiple of 26 leaves the letter unchanged.
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