AMC 10 · 2023 · #5
Grade 6 arithmeticProblem
Maddy and Lara see a list of numbers written on a blackboard. Maddy adds 3 to each number in the list and finds that the sum of her new numbers is 45. Lara multiplies each number in the list by 3 and finds that the sum of her new numbers is also 45. How many numbers are written on the blackboard?
Pick an answer.
AMC 10 2023 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 numbers has sum $S$ and length $n$. Adding $3$ to each entry produces a list whose sum is $45$ (Maddy). Multiplying each entry by $3$ also produces a list whose sum is $45$ (Lara). Find $n$.
Givens: Let $S$ = sum of the original numbers, $n$ = how many numbers there are; Maddy's sum: $S + 3n = 45$; Lara's sum: $3S = 45$; Answer choices: (A) $10$, (B) $5$, (C) $6$, (D) $8$, (E) $9$
Unknowns: $n$ — the number of numbers on the blackboard
Understand
Restated: A list of numbers has sum $S$ and length $n$. Adding $3$ to each entry produces a list whose sum is $45$ (Maddy). Multiplying each entry by $3$ also produces a list whose sum is $45$ (Lara). Find $n$.
Givens: Let $S$ = sum of the original numbers, $n$ = how many numbers there are; Maddy's sum: $S + 3n = 45$; Lara's sum: $3S = 45$; Answer choices: (A) $10$, (B) $5$, (C) $6$, (D) $8$, (E) $9$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #11 Work Backwards, #9 Solve an Easier Related Problem
The problem packs two facts into one paragraph; Tool #7 (Subproblems) splits them: (i) Lara's clue alone tells us the original sum $S$, and (ii) Maddy's clue then tells us how many numbers added up to give that increase. Tool #11 (Work Backwards) applies in step (i) — Lara's sum was *tripled* to reach $45$, so undo the triple to recover the original. A Tool #9 (Easier Problem) sanity check — "what if the list were $\{5,5,5\}$?" — reassures that the structure is right. Full algebra (#13) would work but is overkill for two one-step inversions.
Execute — Answer: A
3.OA.B.5 Step 1 - Use Lara's clue alone.
- Multiplying every number by $3$ multiplies the total by $3$.
- So $3 \times (\text{original sum}) = 45$.
- Work backwards by dividing by $3$.
💡 Lara's $3$-times rule is just one undo away from the original — Grade 3 multiplication/division reverses cleanly.
6.EE.A.2 Step 2 - Now use Maddy's clue.
- Adding $3$ to each of the $n$ numbers adds $3n$ to the total.
- So Maddy's new sum is $S + 3n$, which equals $45$.
💡 "$3$ added to each of $n$ numbers" totals $3n$ extra — a Grade 6 expression that records the bookkeeping.
6.EE.B.7 Step 3 Substitute $S = 15$ from step 1 and isolate $3n$ by subtracting.
💡 Once we know the original sum, the extra $30$ in Maddy's pile is pure $3$-per-number bonus.
3.OA.B.6 Step 4 - Divide by $3$ to find $n$.
- So there are $10$ numbers on the blackboard, matching choice (A).
💡 If $30$ extra came from $+3$ per number, then $30 \div 3 = 10$ numbers — Grade 3 division-as-unknown-factor.
3.OA.B.5 Use Lara's clue alone. Multiplying every number by $3$ multiplies the total by $ 6.EE.A.2 Now use Maddy's clue. Adding $3$ to each of the $n$ numbers adds $3n$ to the tot 6.EE.B.7 Substitute $S = 15$ from step 1 and isolate $3n$ by subtracting. 3.OA.B.6 Divide by $3$ to find $n$. So there are $10$ numbers on the blackboard, matching Review
Reasonableness: Plug back. Pick any list of $10$ numbers summing to $15$ — for example all $1.5$s. Lara multiplies each by $3$ to get ten $4.5$s, summing to $45$. Maddy adds $3$ to each $1.5$ to get ten $4.5$s, summing to $45$. Both clues check out, and (A) $n = 10$ is consistent with both equations.
Alternative: Tool #6 (Guess and Check). The Lara equation $3S = 45$ forces $S = 15$ regardless of $n$. For each candidate $n \in \{5, 6, 8, 9, 10\}$, check $S + 3n = 15 + 3n \stackrel{?}{=} 45$: $n=5$ gives $30$, $n=6$ gives $33$, $n=8$ gives $39$, $n=9$ gives $42$, $n=10$ gives $45$ ✓ — answer (A).
CCSS standards used (min grade 6)
3.OA.B.5Apply properties of operations as strategies to multiply and divide (Recognizing that scaling every number by $3$ scales the total by $3$, so the original sum is one division away.)3.OA.B.6Understand division as an unknown-factor problem (Inverting "$3n = 30$" via $n = 30 \div 3 = 10$.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Writing Maddy's sum as $S + 3n$ — a single expression that captures "add $3$ to each of $n$ items".)6.EE.B.7Solve real-world problems by writing and solving equations of the form $px = q$ (Solving $15 + 3n = 45$ for $n$ to land on $n = 10$.)
⭐ This AMC 10 problem only needs Grade 6 expression-and-equation thinking — Lara's clue pins the original sum at $15$, and Maddy's extra $30$ is just $+3$ per number, so there are $10$ numbers.
⭐ This AMC 10 problem only needs Grade 6 expression-and-equation thinking — Lara's clue pins the original sum at $15$, and Maddy's extra $30$ is just $+3$ per number, so there are $10$ numbers.