AMC 10 · 2023 · #23
Grade 8 arithmeticProblem
If the positive integer n has positive integer divisors a and b with n=ab, then a and b are said to be 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?
Pick an answer.
AMC 10 2023 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: 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 $N$.
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) $19$
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
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., $a^2 + 20a = b^2 + 23b$.
💡 Grade 6 "write expressions to solve problems" — each factor-pair clue is one product expression.
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 $(2a + 20)^2 - 400 = (2b + 23)^2 - 529$.
💡 Grade 8 "use square symbols" — repackage the linear parts as squared binomials so a difference of squares appears.
8.EE.A.2 Step 3 - Rearrange to isolate the difference of squares.
- From step 2, $(2b+23)^2 - (2a+20)^2 = 529 - 400 = 129$.
💡 Subtract sides; we have one squared expression minus another equaling a small constant.
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 $(2b - 2a + 3)(2b + 2a + 43) = 129$.
💡 The classic $X^2 - Y^2 = (X-Y)(X+Y)$ factoring trick collapses a tough equation into two factors.
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 $2b+2a+43 = 43$.
💡 Grade 6 "find factor pairs" — only two pairs to try.
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 $N = 924$.
💡 Grade 8 one-variable linear solve — substitute and unwind to get $b$, then $a$.
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 $N = 924$.
💡 Second case forces $a=0$, which is not a positive divisor — case eliminated.
2.NBT.A.1 Step 8 - Compute the sum of digits of $N = 924$.
- $9 + 2 + 4 = 15$, which is choice (C).
💡 Grade 2 place-value step — add the three digits of $924$.
6.EE.B.6 Translate the two clues into algebra. Let one factor be $a$, so its partner is $ 8.EE.A.2 Turn each side into a perfect square so the equation has the form (square) $-$ ( 8.EE.A.2 Rearrange to isolate the difference of squares. From step 2, $(2b+23)^2 - (2a+20 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 6.NS.B.4 List the positive factor pairs of $129$. Since $129 = 3 \cdot 43$ with $3$ and $ 8.EE.C.7 Solve system (i): from $2b - 2a + 3 = 1$ get $b - a = -1$, so $a = b + 1$. Subst 8.EE.C.7 Solve system (ii): $2b - 2a + 3 = 3$ gives $b = a$, and $2b + 2a + 43 = 43$ give 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.1Understand 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.6Use 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.4Find 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.2Use 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.7Solve 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 $15$.
⭐ 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 $15$.