AMC 10 · 2023 · #23

Grade 8 arithmetic

factorsdivisibility-rulesdigit-sumdifference-of-squares convert-to-algebraidentify-subproblemssystematic-enumeration ↑ Prerequisites: factorsdivisibility-rules

📏 Long solution 💡 3 insights

Problem

If the positive integer $n$ has positive integer divisors $a$ and $b$ with $n = ab$ , then $a$ and $b$ are said to be $\textit{complementary}$ divisors of $n$ . Suppose that $N$ is a positive integer that has one complementary pair of divisors that differ by $20$ and another pair of complementary divisors that differ by $23$ . What is the sum of the digits of $N$ ?

$\textbf{(A) } 9 \qquad \textbf{(B) } 13\qquad \textbf{(C) } 15 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: A positive integer $N$ has two factor pairs of the form $N = ab$. In one pair the two factors differ by $20$, and in another pair they differ by $23$. Find the sum of the digits of $N$.

Givens: $N$ is a positive integer; Some factor pair $(a, a+20)$ of $N$ satisfies $a(a+20) = N$; Some factor pair $(b, b+23)$ of $N$ satisfies $b(b+23) = N$; $a$ and $b$ are positive integers; Answer choices: (A) $9$, (B) $13$, (C) $15$, (D) $17$, (E) $19$

Unknowns: $N$ and the sum of its digits

Understand

Restated: A positive integer $N$ has two factor pairs of the form $N = ab$. In one pair the two factors differ by $20$, and in another pair they differ by $23$. Find the sum of the digits of $N$.

Plan

Primary tool: #13 Convert to Algebra

Secondary: #7 Identify Subproblems, #2 Make a Systematic List

Two factor-pair conditions about a single unknown $N$ scream Tool #13 (Convert to Algebra) — write $N = a(a+20) = b(b+23)$ and equate. Tool #7 (Identify Subproblems) breaks the algebra into a clean chain: (i) match the two products, (ii) rearrange to a difference of squares, (iii) factor the constant on the right side. Tool #2 (Make a Systematic List) finishes by listing the factor pairs of that small constant ($129 = 3 \cdot 43$) — there are only a couple, so each gives one quick mini-system to solve.

Execute — Answer: C

#13 Convert to Algebra 6.EE.B.6 Step 1

Translate the two clues into algebra.
Let one factor be $a$, so its partner is $a+20$, and $N = a(a+20)$.
Similarly let $b$ and $b+23$ be the other pair, so $N = b(b+23)$.
Set them equal: $a(a+20) = b(b+23)$, i.e., $a^2 + 20a = b^2 + 23b$.

$$a^2 + 20a = b^2 + 23b$$

💡 Grade 6 "write expressions to solve problems" — each factor-pair clue is one product expression.

#13 Convert to Algebra 8.EE.A.2 Step 2

Turn each side into a perfect square so the equation has the form (square) $-$ (square) $=$ constant.
Multiply everything by $4$ so the linear terms become $\pm 2\cdot 2\cdot 10 = 40$ and $\pm 2\cdot 2\cdot 11.5 = 46$ for $a$ and $b$ respectively, allowing clean completion of squares.
Then $4a^2 + 80a = 4b^2 + 92b$ becomes $(2a + 20)^2 - 400 = (2b + 23)^2 - 529$.

$$(2a+20)^2 - 400 = (2b+23)^2 - 529$$

💡 Grade 8 "use square symbols" — repackage the linear parts as squared binomials so a difference of squares appears.

#7 Identify Subproblems 8.EE.A.2 Step 3

Rearrange to isolate the difference of squares.
From step 2, $(2b+23)^2 - (2a+20)^2 = 529 - 400 = 129$.

$$(2b+23)^2 - (2a+20)^2 = 129$$

💡 Subtract sides; we have one squared expression minus another equaling a small constant.

#13 Convert to Algebra 8.EE.A.2 Step 4

Factor the left as $X^2 - Y^2 = (X-Y)(X+Y)$ with $X = 2b+23$ and $Y = 2a+20$.
So $(2b - 2a + 3)(2b + 2a + 43) = 129$.

$$(2b - 2a + 3)(2b + 2a + 43) = 129$$

💡 The classic $X^2 - Y^2 = (X-Y)(X+Y)$ factoring trick collapses a tough equation into two factors.

#2 Make a Systematic List 6.NS.B.4 Step 5

List the positive factor pairs of $129$.
Since $129 = 3 \cdot 43$ with $3$ and $43$ both prime, the only positive factorizations are $1 \cdot 129$ and $3 \cdot 43$.
For positive $a, b$ the right factor $2b + 2a + 43$ is the larger one, so we get two systems: (i) $2b-2a+3 = 1$ and $2b+2a+43 = 129$; (ii) $2b-2a+3 = 3$ and $2b+2a+43 = 43$.

$$129 = 1 \cdot 129 = 3 \cdot 43$$

💡 Grade 6 "find factor pairs" — only two pairs to try.

#13 Convert to Algebra 8.EE.C.7 Step 6

Solve system (i): from $2b - 2a + 3 = 1$ get $b - a = -1$, so $a = b + 1$.
Substitute into $2b + 2a + 43 = 129$: $2b + 2(b+1) + 43 = 129$, i.e., $4b + 45 = 129$, so $4b = 84$ and $b = 21$.
Then $a = 22$.
Check: $N = 22 \cdot 42 = 924$ and $N = 21 \cdot 44 = 924$.
Both products match, so $N = 924$.

$$b = 21,\;a = 22,\;N = 22\cdot 42 = 21\cdot 44 = 924$$

💡 Grade 8 one-variable linear solve — substitute and unwind to get $b$, then $a$.

#13 Convert to Algebra 8.EE.C.7 Step 7

Solve system (ii): $2b - 2a + 3 = 3$ gives $b = a$, and $2b + 2a + 43 = 43$ gives $4a = 0$, so $a = 0$.
But $a$ must be a positive integer (it is a divisor), so this case is rejected.
Therefore the unique answer is $N = 924$.

$$a = 0 \text{ rejected; only } N = 924 \text{ works}$$

💡 Second case forces $a=0$, which is not a positive divisor — case eliminated.

#7 Identify Subproblems 2.NBT.A.1 Step 8

Compute the sum of digits of $N = 924$.
$9 + 2 + 4 = 15$, which is choice (C).

$$9 + 2 + 4 = 15 \;\Rightarrow\; \textbf{(C)}$$

💡 Grade 2 place-value step — add the three digits of $924$.

[1] #13 6.EE.B.6 Translate the two clues into algebra. Let one factor be $a$, so its partner is $

[2] #13 8.EE.A.2 Turn each side into a perfect square so the equation has the form (square) $-$ (

[3] #7 8.EE.A.2 Rearrange to isolate the difference of squares. From step 2, $(2b+23)^2 - (2a+20

[4] #13 8.EE.A.2 Factor the left as $X^2 - Y^2 = (X-Y)(X+Y)$ with $X = 2b+23$ and $Y = 2a+20$. So

[5] #2 6.NS.B.4 List the positive factor pairs of $129$. Since $129 = 3 \cdot 43$ with $3$ and $

[6] #13 8.EE.C.7 Solve system (i): from $2b - 2a + 3 = 1$ get $b - a = -1$, so $a = b + 1$. Subst

[7] #13 8.EE.C.7 Solve system (ii): $2b - 2a + 3 = 3$ gives $b = a$, and $2b + 2a + 43 = 43$ give

[8] #7 2.NBT.A.1 Compute the sum of digits of $N = 924$. $9 + 2 + 4 = 15$, which is choice (C).

Review

Reasonableness: Verify the factor pairs of $N = 924$ directly. $924 = 21 \cdot 44$ (difference $44 - 21 = 23$ \checkmark) and $924 = 22 \cdot 42$ (difference $42 - 22 = 20$ \checkmark). Both required gaps appear in the factor list. The factorization $924 = 2^2 \cdot 3 \cdot 7 \cdot 11$ has $(2+1)(1+1)(1+1)(1+1) = 24$ divisors, so plenty of factor pairs exist — and both required gaps happen to be among them. The digit sum $9 + 2 + 4 = 15$ matches exactly one answer choice.

Alternative: Tool #6 (Guess and Check) directly on the answer choices: each choice is a digit sum, so plausible $N$'s are small multiples of the digit-sum class. Or one can search small $a$: for each $a$ compute $N = a(a+20)$ and ask whether $N$ also has a factor pair differing by $23$. The first hit is $a = 22, N = 924$, and the digit sum is $15$.

CCSS standards used (min grade 8)

2.NBT.A.1 Understand that the three digits of a three-digit number represent amounts of hundreds, tens, and ones (Adding the digits $9 + 2 + 4$ of $N = 924$ at the very end.)
6.EE.B.6 Use variables to represent numbers and write expressions when solving a real-world or mathematical problem (Naming the unknown factors $a$ and $b$ and writing $N = a(a+20) = b(b+23)$.)
6.NS.B.4 Find the greatest common factor of two whole numbers and the least common multiple of two whole numbers; recognize when one is a factor (Listing the positive factor pairs of $129 = 3 \cdot 43$ to narrow the systems to two cases.)
8.EE.A.2 Use square root and cube root symbols to represent solutions to equations of the form $x^2 = p$ and $x^3 = p$ (Rewriting $a^2 + 20a$ as $(2a+20)^2/4 - 100$ etc. and factoring $X^2 - Y^2$ as $(X-Y)(X+Y)$.)
8.EE.C.7 Solve linear equations in one variable (Solving the two small linear systems from the factorization to get $b = 21$, $a = 22$ and $N = 924$.)

⭐ This AMC 10 problem only needs Grade 8 algebra you already know — write the two clues as $N = a(a+20) = b(b+23)$, rearrange to a difference of squares equaling $129$, list its tiny factor pairs, and the only positive solution gives $N = 924$ with digit sum $15$.