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!"?}
Try it yourself first — the explanation is most useful after you’ve attempted it.
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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
#16 Change Focus / Count the Complement 4.OA.A.3Step 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 Make a Systematic List 2.OA.B.2Step 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.
$$1 + 2 + 1 + 4 + 2 + 2 = 12$$
💡 Sweep left to right keeping a single running count so no s is missed or double-counted.
#5 Look for a Pattern 4.OA.C.5Step 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.
💡 Every full lap of 26 letters brings you right back home, so only the leftover part changes the letter.
[1]
#16 4.OA.A.3The shift on any letter is set entirely by how many times that same letter has a
[2]
#2 2.OA.B.2Go through the message and tally every s (capital or small letters are the same
[3]
#5 4.OA.C.5The 12th time a letter appears, it moves 1+2+3+...+12 places to the right. This
[4]
#5 4.NBT.B.6Because 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.3 Solve 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.2 Fluently add and subtract within 20 using mental strategies (Tallying the message to count that s appears 12 times in all.)
4.OA.C.5 Generate 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.6 Find 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.