AMC 8 · 2022 · #6
Easy mode Grade 3Problem
Picture three dots on a number line. The dots are evenly spaced, so the gap between the first and second dot is the same as the gap between the second and third dot.
Each dot stands for a whole number. The middle number is 15. The largest number is 4 times the smallest number.
What is the smallest number?
Pick an answer.
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.
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
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) $8$
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
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.
💡 Equal gaps is a one-line rule — once you see it, the rest is just arithmetic checking on each choice.
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.
💡 Multiplying $4 \times 4$ and subtracting are Grade 3 multiplication / word-problem skills.
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.
💡 Same Grade 3 multiply-and-subtract check, just on the next candidate.
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!
💡 When both differences come out the same, the "equally spaced" rule is satisfied — we've found the smallest number.
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) $6$.
💡 Eliminating the rest of the choices makes sure (C) is the only valid answer — no second contestant slipped through.
3.OA.D.8 Understand what "equally spaced" demands. If the three numbers in order are smal 3.OA.A.3 Try choice (A) $x = 4$. Then largest $= 4 \times 4 = 16$. The gap from $4$ to $1 3.OA.A.3 Try choice (B) $x = 5$. Then largest $= 4 \times 5 = 20$. The gap from $5$ to $1 3.OA.A.3 Try choice (C) $x = 6$. Then largest $= 4 \times 6 = 24$. The gap from $6$ to $1 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.3Solve 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.8Solve 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!
⭐ 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!
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