AMC 8 · 2022 · #6

Grade 3 arithmeticalgebra

Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Direct Ratio Scale-Up

sequences-arithmeticequal-spacinglinear-equations-one-var convert-to-algebra ↑ Prerequisites: linear-equations-one-varsequences-arithmetic

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Problem

Three positive integers are equally spaced on a number line. The middle number is $15,$ and the largest number is $4$ times the smallest number. What is the smallest of these three numbers?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2022 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.

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Toolkit + CCSS Solution

Understand

Restated: Three positive whole numbers sit on a number line with the SAME gap between each pair. The middle number is $15$, and the biggest number is $4$ times the smallest number. What is the smallest number?

Givens: Three positive integers placed equally spaced on a number line; The middle number is $15$; The largest number is $4$ times the smallest number; Answer choices: (A) $4$, (B) $5$, (C) $6$, (D) $7$, (E) $8$

Unknowns: The smallest of the three numbers

Understand

Plan

Primary tool: #6 Guess and Check

Secondary: #3 Eliminate Possibilities

This is a multiple-choice problem where each candidate for the smallest number is a small whole number ($4, 5, 6, 7, 8$). For each candidate $x$, we can compute the largest number ($4x$) and then check whether $15$ sits exactly in the middle of $x$ and $4x$ (equal gaps). Tool #6 (Guess and Check) tests each candidate directly, and Tool #3 (Eliminate Possibilities) lets us cross off the ones that fail. This avoids reaching for algebra (Tool #13) when the bounded answer set makes plug-and-check the fastest, most age-appropriate path.

Execute — Answer: C

#6 Guess and Check 3.OA.D.8 Step 1

Understand what "equally spaced" demands.
If the three numbers in order are smallest, middle, largest, then the gap (middle $-$ smallest) must equal the gap (largest $-$ middle).
That means the middle number is exactly halfway between the smallest and the largest.

$$\text{middle} - \text{smallest} = \text{largest} - \text{middle}$$

💡 Equal gaps is a one-line rule — once you see it, the rest is just arithmetic checking on each choice.

#3 Eliminate Possibilities 3.OA.A.3 Step 2

Try choice (A) $x = 4$.
Then largest $= 4 \times 4 = 16$.
The gap from $4$ to $15$ is $11$, and from $15$ to $16$ is $1$.
The gaps don't match, so (A) is out.

$$15 - 4 = 11,\quad 16 - 15 = 1\;\;\text{(unequal)} \Rightarrow \text{eliminate (A)}$$

💡 Multiplying $4 \times 4$ and subtracting are Grade 3 multiplication / word-problem skills.

#3 Eliminate Possibilities 3.OA.A.3 Step 3

Try choice (B) $x = 5$.
Then largest $= 4 \times 5 = 20$.
The gap from $5$ to $15$ is $10$, but from $15$ to $20$ is $5$.
Gaps unequal, so (B) is out.

$$15 - 5 = 10,\quad 20 - 15 = 5\;\;\text{(unequal)} \Rightarrow \text{eliminate (B)}$$

💡 Same Grade 3 multiply-and-subtract check, just on the next candidate.

#6 Guess and Check 3.OA.A.3 Step 4

Try choice (C) $x = 6$.
Then largest $= 4 \times 6 = 24$.
The gap from $6$ to $15$ is $9$, and from $15$ to $24$ is also $9$.
Gaps are equal — this works!

$$15 - 6 = 9,\quad 24 - 15 = 9\;\;\checkmark$$

💡 When both differences come out the same, the "equally spaced" rule is satisfied — we've found the smallest number.

#3 Eliminate Possibilities 3.OA.D.8 Step 5

Confirm by checking the remaining candidates fail.
For $x = 7$: largest $= 28$, gaps $8$ and $13$ — unequal.
For $x = 8$: largest $= 32$, gaps $7$ and $17$ — unequal.
Only (C) survives, so the answer is (C) $6$.

$$x = 7: 15-7=8,\;28-15=13;\;\; x = 8: 15-8=7,\;32-15=17\;\Rightarrow\; \textbf{(C)}\; 6$$

💡 Eliminating the rest of the choices makes sure (C) is the only valid answer — no second contestant slipped through.

[1] #6 3.OA.D.8 Understand what "equally spaced" demands. If the three numbers in order are smal

[2] #3 3.OA.A.3 Try choice (A) $x = 4$. Then largest $= 4 \times 4 = 16$. The gap from $4$ to $1

[3] #3 3.OA.A.3 Try choice (B) $x = 5$. Then largest $= 4 \times 5 = 20$. The gap from $5$ to $1

[4] #6 3.OA.A.3 Try choice (C) $x = 6$. Then largest $= 4 \times 6 = 24$. The gap from $6$ to $1

[5] #3 3.OA.D.8 Confirm by checking the remaining candidates fail. For $x = 7$: largest $= 28$,

Review

Reasonableness: The three numbers $6, 15, 24$ are positive integers in increasing order, the middle is $15$, the largest $24 = 4 \times 6$ is exactly four times the smallest, and the gaps $15 - 6 = 9$ and $24 - 15 = 9$ are equal. Every condition checks out, and $6$ appears in the answer list as choice (C).

Alternative: Tool #13 (Convert to Algebra) lets the smallest be $x$, the largest $4x$, and uses the middle-is-halfway property: $\dfrac{x + 4x}{2} = 15 \Rightarrow \dfrac{5x}{2} = 15 \Rightarrow x = 6$. It reaches the same answer but takes one more abstraction step than just plugging in the five small choices.

CCSS standards used (min grade 3)

3.OA.A.3 Solve multiplication and division word problems within 100 (Computing the largest number $4 \times x$ for each candidate ($4 \times 4, 4 \times 5, \dots, 4 \times 8$) — all within 100.)
3.OA.D.8 Solve two-step word problems using four operations within 100 (Checking the "equally spaced" rule by computing two subtractions per candidate and comparing whether the two gaps are equal.)

⭐ This AMC 8 problem only needs Grade 3 multiplication and subtraction you already know — just try each answer choice and see which one makes equal gaps!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 3)

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