AMC 8 · 2014 · #22

Grade 3 algebra

Archetype Arithmetic Expression Decomposition

place-valuelinear-equations-one-vardigit-decomposition convert-to-algebradigit-constraints ↑ Prerequisites: place-valuelinear-equations-one-var

📏 Short solution 💡 3 insights

Problem

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?

$\textbf{(A) }1\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }7\qquad \textbf{(E) }9$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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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.

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

#6 Guess and Check 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).

$$21 \neq (2 \times 1) + (2 + 1) = 5$$

💡 Multiplying and adding single-digit numbers is a Grade 3 fluency skill.

#3 Eliminate Possibilities 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).

$$23 \neq (2 \times 3) + (2 + 3) = 11$$

💡 Each failed check rules out one answer choice — the classic Tool #3 (Eliminate) move on a multiple-choice problem.

#5 Look for a Pattern 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.

$$\text{gap} = (10a + b) - [(a \times b) + (a + b)] = a \times (9 - b)$$

💡 Spotting the common factor $a$ in the gap is a Grade 3 distributive-property observation, not full algebra.

#3 Eliminate Possibilities 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).

$$25 \neq 17, \quad 27 \neq 23$$

💡 The gap formula predicts $2 \times (9-5) = 8$ and $2 \times (9-7) = 4$, matching $25 - 17$ and $27 - 23$ — the pattern holds.

#6 Guess and Check 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 $9 \Rightarrow \textbf{(E)}$.

$$29 = (2 \times 9) + (2 + 9), \quad 59 = (5 \times 9) + (5 + 9)$$

💡 Verifying with two different tens digits confirms $b = 9$ works for every valid $a$ — Grade 3 multi-step word-problem checking.

[1] #6 3.OA.C.7 Test units digit $b = 1$ with tens digit $a = 2$, so the number is $21$. Compute

[2] #3 3.OA.C.7 Test units digit $b = 3$ with $a = 2$, number $= 23$. Compute $(2 \times 3) + (2

[3] #5 3.OA.B.5 Notice a pattern (Tool #5): the right side $(a \times b) + (a + b)$ keeps coming

[4] #3 3.OA.C.7 Quick checks of the remaining wrong choices. Test $b = 5$, $a = 2$: $(2 \times 5

[5] #6 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.7 Fluently 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.5 Apply 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.8 Solve 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 3)

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