AMC 10 · 2002 · #23
Grade 6 arithmeticLet {ak} be a sequence of integers such that a1=1 and am+n=am+an+mn, for all positive integers m and n. Then a12 is
Pick an answer.
AMC 10 2002 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 list of whole numbers starts with $a_1 = 1$, and it obeys the rule $a_{m+n} = a_m + a_n + mn$ for every choice of positive whole numbers $m$ and $n$. Find the twelfth number, $a_{12}$.
Givens: The first term is $a_1 = 1$; For all positive integers $m$ and $n$, $a_{m+n} = a_m + a_n + mn$; The terms are integers; Answer choices: (A) $45$, (B) $56$, (C) $67$, (D) $78$, (E) $89$
Unknowns: The value of the twelfth term $a_{12}$
Understand
Restated: A list of whole numbers starts with $a_1 = 1$, and it obeys the rule $a_{m+n} = a_m + a_n + mn$ for every choice of positive whole numbers $m$ and $n$. Find the twelfth number, $a_{12}$.
Givens: The first term is $a_1 = 1$; For all positive integers $m$ and $n$, $a_{m+n} = a_m + a_n + mn$; The terms are integers; Answer choices: (A) $45$, (B) $56$, (C) $67$, (D) $78$, (E) $89$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #9 Solve an Easier Related Problem, #4 Introduce a Variable
The rule $a_{m+n}=a_m+a_n+mn$ has two free letters, which looks heavy. Tool #9 (Solve an Easier Related Problem) tames it: since the rule holds for every $m$ and $n$, set the easiest value $n=1$, turning the two-letter rule into a plain step-by-step recipe for getting the next term. Tool #5 (Look for a Pattern) then does the real work — list the first few terms the recipe produces and recognize them as the familiar triangular numbers $1,3,6,10,\dots$, whose $k$th value is $1+2+\cdots+k$. Tool #4 (Introduce a Variable) keeps the bookkeeping honest by tracking the index $m$ as we climb from $a_1$ up to $a_{12}$.
Execute — Answer: D
6.EE.A.2 Step 1 Set $n=1$ to get a step rule
- The rule is promised for every $m$ and $n$, so choose the simplest partner: let $n=1$.
- Then $a_{m+1} = a_m + a_1 + m\cdot 1$.
- Since $a_1 = 1$, this becomes $a_{m+1} = a_m + 1 + m = a_m + (m+1)$.
- In words: to jump from the $m$th term to the next one, add $m+1$.
- The scary two-letter rule has become a simple 'add the next counting number' recipe.
💡 Because the rule works for all inputs, we may plug in the easiest one, $n=1$, and read off how each term grows from the last.
4.OA.C.5 Step 2 Build the terms and spot the pattern
- Start from $a_1 = 1$ and apply the recipe, adding $m+1$ each step.
- $a_2 = a_1 + 2 = 3$; $a_3 = a_2 + 3 = 6$; $a_4 = a_3 + 4 = 10$; $a_5 = a_4 + 5 = 15$.
- The list $1, 3, 6, 10, 15, \dots$ is the triangular numbers: each term is a running total of the counting numbers, so $a_k = 1 + 2 + 3 + \cdots + k$.
💡 Adding $2$, then $3$, then $4$ each step piles the counting numbers on top of each other, which is exactly what a triangular number is.
4.OA.A.3 Step 3 Add up to the twelfth term
- Using the pattern, $a_{12} = 1 + 2 + 3 + \cdots + 12$.
- Add it fast by pairing ends: $1+12 = 13$, $2+11 = 13$, and so on, giving $6$ pairs that each total $13$.
- So $a_{12} = 6 \times 13 = 78$.
- That is choice (D).
💡 Pairing the smallest and largest leftovers turns a long sum into a few equal chunks you can multiply.
6.EE.A.2 The rule is promised for every $m$ and $n$, so choose the simplest partner: let 4.OA.C.5 Start from $a_1 = 1$ and apply the recipe, adding $m+1$ each step. $a_2 = a_1 + 4.OA.A.3 Using the pattern, $a_{12} = 1 + 2 + 3 + \cdots + 12$. Add it fast by pairing en Review
Reasonableness: Check $78$ against the original two-letter rule without the $n=1$ shortcut. Take $m=n=6$: $a_{12} = a_6 + a_6 + 6\cdot 6 = 2a_6 + 36$. From the list $a_6 = 21$, so $a_{12} = 2\cdot 21 + 36 = 42 + 36 = 78$ — the same answer by a completely different split, which is strong confirmation. The value is also sensible in size: $a_{12}$ should be near $1+2+\cdots+12$, comfortably between the small decoy $45$ and the large decoy $89$.
Alternative: Recognize the closed form directly. The triangular pattern gives $a_k = \tfrac{k(k+1)}{2}$, so $a_{12} = \tfrac{12\cdot 13}{2} = \tfrac{156}{2} = 78$, again choice (D). Notice the five answer choices sit $11$ apart ($45, 56, 67, 78, 89$); the formula lands squarely on $78$, so no near-miss choice can compete.
CCSS standards used (min grade 6)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Substituting $n=1$ into $a_{m+n}=a_m+a_n+mn$ to rewrite it as the step rule $a_{m+1}=a_m+(m+1)$.)4.OA.C.5Generate a number or shape pattern following a given rule (Generating $1,3,6,10,15,\dots$ from the step rule and recognizing them as the triangular numbers.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Adding $1+2+\cdots+12$ by pairing ends to get $6\times 13 = 78$.)
⭐ When a rule works for every input, plug in the easiest one — here $n=1$ turns the rule into 'add the next number,' building the triangular numbers straight up to $78$.
⭐ When a rule works for every input, plug in the easiest one — here $n=1$ turns the rule into 'add the next number,' building the triangular numbers straight up to $78$.
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