AMC 8 · 2014 · #22
Grade 3 algebraProblem
A 2-digit number is such that the product of the digits plus the sum of the digits is equal to the number. What is the units digit of the number?
Pick an answer.
AMC 8 2014 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: Find a two-digit number where (product of its two digits) $+$ (sum of its two digits) equals the number itself. The question asks for the units digit of any such number.
Givens: The number has exactly $2$ digits, so it is between $10$ and $99$; Let the tens digit be $a$ and the units digit be $b$; The number's value is $10a + b$; Product of digits $= a \times b$; Sum of digits $= a + b$; Condition: $10a + b = (a \times b) + (a + b)$; Answer choices: (A) $1$, (B) $3$, (C) $5$, (D) $7$, (E) $9$
Unknowns: The units digit $b$ of every two-digit number that satisfies the condition
Understand
Restated: Find a two-digit number where (product of its two digits) $+$ (sum of its two digits) equals the number itself. The question asks for the units digit of any such number.
Givens: The number has exactly $2$ digits, so it is between $10$ and $99$; Let the tens digit be $a$ and the units digit be $b$; The number's value is $10a + b$; Product of digits $= a \times b$; Sum of digits $= a + b$; Condition: $10a + b = (a \times b) + (a + b)$; Answer choices: (A) $1$, (B) $3$, (C) $5$, (D) $7$, (E) $9$
Plan
Primary tool: #6 Guess and Check
Secondary: #3 Eliminate Possibilities, #5 Look for a Pattern
The answer choices give only $5$ possible units digits ($1, 3, 5, 7, 9$), so we can test each one directly. For each candidate units digit $b$, pick a simple tens digit (say $a = 2$) and check whether $10a + b$ equals $a \times b + a + b$. Tool #6 (Guess and Check) does the testing; Tool #3 (Eliminate) discards every candidate that fails. After two or three checks a clear pattern jumps out (Tool #5), confirming the winning digit without needing algebra.
Execute — Answer: E
3.OA.C.7 Step 1 - Test units digit $b = 1$ with tens digit $a = 2$, so the number is $21$.
- Compute (product) $+$ (sum) $= (2 \times 1) + (2 + 1) = 2 + 3 = 5$.
- That is not $21$, so $b = 1$ fails.
- Eliminate (A).
💡 Multiplying and adding single-digit numbers is a Grade 3 fluency skill.
3.OA.C.7 Step 2 - Test units digit $b = 3$ with $a = 2$, number $= 23$.
- Compute $(2 \times 3) + (2 + 3) = 6 + 5 = 11$.
- Not $23$, so eliminate (B).
💡 Each failed check rules out one answer choice — the classic Tool #3 (Eliminate) move on a multiple-choice problem.
3.OA.B.5 Step 3 - Notice a pattern (Tool #5): the right side $(a \times b) + (a + b)$ keeps coming out smaller than the number $10a + b$, and the gap is exactly $10a - (a \times b + a) = 9a - a \times b = a \times (9 - b)$.
- The gap shrinks to $0$ only when $9 - b = 0$, i.e., $b = 9$.
- Let's verify with $b = 5$ and $b = 7$ first to be sure.
💡 Spotting the common factor $a$ in the gap is a Grade 3 distributive-property observation, not full algebra.
3.OA.C.7 Step 4 - Quick checks of the remaining wrong choices.
- Test $b = 5$, $a = 2$: $(2 \times 5) + (2 + 5) = 10 + 7 = 17 \neq 25$.
- Test $b = 7$, $a = 2$: $(2 \times 7) + (2 + 7) = 14 + 9 = 23 \neq 27$.
- Both fail — eliminate (C) and (D).
💡 The gap formula predicts $2 \times (9-5) = 8$ and $2 \times (9-7) = 4$, matching $25 - 17$ and $27 - 23$ — the pattern holds.
3.OA.D.8 Step 5 - Test $b = 9$ with $a = 2$: number $= 29$.
- Compute $(2 \times 9) + (2 + 9) = 18 + 11 = 29$.
- It works!
- Try another tens digit to be sure: $a = 5$ gives number $= 59$ and $(5 \times 9) + (5 + 9) = 45 + 14 = 59$.
- Also works.
- So the units digit must be $9 \Rightarrow \textbf{(E)}$.
💡 Verifying with two different tens digits confirms $b = 9$ works for every valid $a$ — Grade 3 multi-step word-problem checking.
3.OA.C.7 Test units digit $b = 1$ with tens digit $a = 2$, so the number is $21$. Compute 3.OA.C.7 Test units digit $b = 3$ with $a = 2$, number $= 23$. Compute $(2 \times 3) + (2 3.OA.B.5 Notice a pattern (Tool #5): the right side $(a \times b) + (a + b)$ keeps coming 3.OA.C.7 Quick checks of the remaining wrong choices. Test $b = 5$, $a = 2$: $(2 \times 5 3.OA.D.8 Test $b = 9$ with $a = 2$: number $= 29$. Compute $(2 \times 9) + (2 + 9) = 18 + Review
Reasonableness: The gap formula $a \times (9 - b)$ explains everything: the number always exceeds the (product $+$ sum) by $a \times (9 - b)$ blocks. That gap is zero only when $b = 9$, and it works for any nonzero tens digit. So $19, 29, 39, \ldots, 99$ all satisfy the condition (e.g. $19 = 1 \cdot 9 + 1 + 9 = 9 + 10$). Every such number ends in $9$ — answer (E) is consistent.
Alternative: Tool #13 (Convert to Algebra) gives a one-line proof: $10a + b = ab + a + b \Rightarrow 9a = ab \Rightarrow b = 9$ (since $a \neq 0$). Faster on paper, but Guess-and-Check $+$ Eliminate is friendlier for younger solvers and uses no algebra.
CCSS standards used (min grade 3)
3.OA.C.7Fluently multiply and divide within $100$ (Computing the product $a \times b$ for each tested digit pair (e.g., $2 \times 9 = 18$, $5 \times 9 = 45$).)3.OA.B.5Apply properties of operations as strategies to multiply and add (Recognizing that the gap $(10a + b) - (ab + a + b)$ factors as $a \times (9 - b)$ — a distributive-property observation.)3.OA.D.8Solve two-step word problems using the four operations (Verifying each candidate by combining a multiplication and an addition, then comparing with the two-digit number.)
⭐ This AMC 8 problem only needs Grade 3 multiplication-and-addition checking — just test each answer choice and the right units digit shows itself!
⭐ This AMC 8 problem only needs Grade 3 multiplication-and-addition checking — just test each answer choice and the right units digit shows itself!
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