AMC 10 · 2012 · #11
Grade 6 arithmeticA dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?
Pick an answer.
AMC 10 2012 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 dessert chef plans one dessert for each of the seven days from Sunday through Saturday. Each day's dessert is cake, pie, ice cream, or pudding, and the same dessert cannot appear two days in a row. Friday must be cake because of a birthday. Count how many different seven-day menus are possible.
Givens: The week runs Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday — $7$ days; Each day's dessert is one of $4$ options: cake, pie, ice cream, or pudding; No dessert may be served on two consecutive days; Friday's dessert must be cake; Answer choices: (A) $729$, (B) $972$, (C) $1024$, (D) $2187$, (E) $2304$
Unknowns: How many different $7$-day dessert menus satisfy all the rules
Understand
Restated: A dessert chef plans one dessert for each of the seven days from Sunday through Saturday. Each day's dessert is cake, pie, ice cream, or pudding, and the same dessert cannot appear two days in a row. Friday must be cake because of a birthday. Count how many different seven-day menus are possible.
Givens: The week runs Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday — $7$ days; Each day's dessert is one of $4$ options: cake, pie, ice cream, or pudding; No dessert may be served on two consecutive days; Friday's dessert must be cake; Answer choices: (A) $729$, (B) $972$, (C) $1024$, (D) $2187$, (E) $2304$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #7 Identify Subproblems, #5 Look for a Pattern
The question asks "how many ways," so Tool #2 (Make a Systematic List) drives the count: decide the desserts in a smart order instead of guessing. Tool #7 (Identify Subproblems) turns one hard "count the whole week" into seven easy "count one day" pieces that multiply together. Tool #5 (Look for a Pattern) spots that every day except Friday has the same $3$ choices, which collapses the product into a single power of $3$ and guards against the trap of also giving Friday a factor.
Execute — Answer: A
3.OA.A.1 Step 1 Anchor Friday, then count outward
- Friday is forced to be cake, so start there and decide the other days in an order that always steps to a day next to one already chosen — for example Thursday, then Wednesday, Tuesday, Monday, Sunday, and separately Saturday.
- Each such day only has to differ from the single neighbor already fixed, so it may be any dessert except that one: $4-1=3$ choices.
💡 Fixing one neighbor rules out exactly one dessert, leaving three.
6.EE.A.1 Step 2 Multiply the independent day-counts
- There are $6$ days other than Friday (Sunday, Monday, Tuesday, Wednesday, Thursday, Saturday), and each independently has $3$ valid choices once its already-decided neighbor is known.
- Friday itself is fixed, contributing a factor of $1$.
- By the multiplication principle the total number of menus is the product of the per-day counts.
💡 Independent choices combine by multiplying, and six identical factors of $3$ pack into $3^6$.
5.NBT.B.5 Step 3 Evaluate the power of 3
- Compute $3^6$ by repeated multiplication: $3^2=9$, $3^3=27$, $3^4=81$, $3^5=243$, $3^6=729$.
- That matches choice (A).
- The value $729$ sits neatly between the answer choices, and the clean power-of-$3$ form confirms the count, so the answer is (A).
💡 Six threes multiplied together is $729$, the smallest listed choice.
3.OA.A.1 Friday is forced to be cake, so start there and decide the other days in an orde 6.EE.A.1 There are $6$ days other than Friday (Sunday, Monday, Tuesday, Wednesday, Thursd 5.NBT.B.5 Compute $3^6$ by repeated multiplication: $3^2=9$, $3^3=27$, $3^4=81$, $3^5=243$ Review
Reasonableness: Check the biggest trap: if you forget that Friday is fixed and give all $7$ days $3$ choices, you get $3^7=2187$, which is exactly the distractor (D). Anchoring Friday removes one factor of $3$, dropping $2187$ to $3^6=729$ — so (A) is right and (D) is the intended mistake. Also $4^{7}$ (ignoring the no-repeat rule) would be far larger than any choice, confirming the constraints really do the pruning.
Alternative: Count every valid week with no birthday rule first: Sunday has $4$ choices and each later day has $3$ (just differ from the day before), giving $4\times 3^{6}$ menus. By symmetry cake is equally likely to land on Friday as any of the $4$ desserts, so exactly one-fourth of these menus have cake on Friday: $\frac{4\times 3^{6}}{4}=3^{6}=729$, choice (A).
CCSS standards used (min grade 6)
3.OA.A.1Interpret products of whole numbers as equal groups (Reading each non-Friday day as one group of $3$ available desserts (four options minus the forbidden neighbor).)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Packing the six independent factors of $3$ into the expression $3^6$ via the multiplication principle.)5.NBT.B.5Fluently multiply multi-digit whole numbers (Evaluating $3^6$ step by step up to $729$ and matching it to a choice.)
⭐ Pin down the forced day first, then every other day has just $3$ choices, so multiply: $3^6=729$.
⭐ Pin down the forced day first, then every other day has just $3$ choices, so multiply: $3^6=729$.
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