AMC 10 · 2023 · #5

Grade 6 arithmetic

linear-equations-one-varmean-median-mode-rangesystems-of-equations identify-subproblemswork-backwardsconvert-to-algebra ↑ Prerequisites: linear-equations-one-var

📏 Short solution 💡 2 insights

Problem

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?

$\textbf{(A) }10\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }8\qquad\textbf{(E) }9$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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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

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

#11 Work Backwards 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 $3$.

$$3S = 45 \;\Rightarrow\; S = \dfrac{45}{3} = 15$$

💡 Lara's $3$-times rule is just one undo away from the original — Grade 3 multiplication/division reverses cleanly.

#7 Identify Subproblems 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 $45$.

$$S + 3n = 45$$

💡 "$3$ added to each of $n$ numbers" totals $3n$ extra — a Grade 6 expression that records the bookkeeping.

#11 Work Backwards 6.EE.B.7 Step 3

Substitute $S = 15$ from step 1 and isolate $3n$ by subtracting.

$$15 + 3n = 45 \;\Rightarrow\; 3n = 30$$

💡 Once we know the original sum, the extra $30$ in Maddy's pile is pure $3$-per-number bonus.

#7 Identify Subproblems 3.OA.B.6 Step 4

Divide by $3$ to find $n$.
So there are $10$ numbers on the blackboard, matching choice (A).

$$n = \dfrac{30}{3} = 10 \;\Rightarrow\; \textbf{(A)}$$

💡 If $30$ extra came from $+3$ per number, then $30 \div 3 = 10$ numbers — Grade 3 division-as-unknown-factor.

[1] #11 3.OA.B.5 Use Lara's clue alone. Multiplying every number by $3$ multiplies the total by $

[2] #7 6.EE.A.2 Now use Maddy's clue. Adding $3$ to each of the $n$ numbers adds $3n$ to the tot

[3] #11 6.EE.B.7 Substitute $S = 15$ from step 1 and isolate $3n$ by subtracting.

[4] #7 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.5 Apply 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.6 Understand division as an unknown-factor problem (Inverting "$3n = 30$" via $n = 30 \div 3 = 10$.)
6.EE.A.2 Write, 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.7 Solve 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.