AMC 10 · 2019 · #6

Grade 6 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

factorialpolynomial-factoringfactorsdigit-sum pattern-recognitionguess-and-check ↑ Prerequisites: factorialpolynomial-factoring

📏 Medium solution 💡 3 insights

Problem

There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$ ?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2019 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.

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Toolkit + CCSS Solution

Understand

Restated: Find a positive integer $n$ for which $(n+1)! + (n+2)! = n! \cdot 440$. Then add the digits of that $n$.

Givens: $(n+1)! + (n+2)! = n! \cdot 440$; $n$ is a positive integer; Choices: (A) $3$, (B) $8$, (C) $10$, (D) $11$, (E) $12$

Unknowns: The value of $n$; The sum of the digits of $n$

Understand

Restated: Find a positive integer $n$ for which $(n+1)! + (n+2)! = n! \cdot 440$. Then add the digits of that $n$.

Givens: $(n+1)! + (n+2)! = n! \cdot 440$; $n$ is a positive integer; Choices: (A) $3$, (B) $8$, (C) $10$, (D) $11$, (E) $12$

Plan

Primary tool: #6 Guess and Check

Secondary: #9 Solve an Easier Related Problem, #5 Look for a Pattern, #3 Eliminate Possibilities

Factorials are scary, but the same $n!$ hides in every term. Divide it out (Tool #9 simplifies the equation), look at the shape that remains (Tool #5: it becomes a product of two consecutive-step integers), then try a few values (Tool #6) and confirm by checking the answer choices' digit sums (Tool #3).

Execute — Answer: C

#9 Solve an Easier Related Problem 5.OA.A.2 Step 1

Notice $(n+1)! = (n+1)\cdot n!$ and $(n+2)! = (n+2)(n+1)\cdot n!$.
So both terms have $n!$ as a common factor.
Pull it out.

$$(n+1)!+(n+2)! = n!\bigl[(n+1)+(n+2)(n+1)\bigr]$$

💡 Factoring out the shared $n!$ turns a factorial puzzle into a small algebra puzzle.

#9 Solve an Easier Related Problem 6.EE.A.3 Step 2

Divide both sides by $n!$ — it cancels cleanly.
What is left is a product of two simple factors.

$$(n+1) + (n+2)(n+1) = 440 \;\Rightarrow\; (n+1)\bigl[1+(n+2)\bigr] = 440 \;\Rightarrow\; (n+1)(n+3) = 440$$

💡 Two whole numbers that differ by $2$ multiply to $440$ — much easier to chase.

#6 Guess and Check 4.OA.B.4 Step 3

Look for two whole numbers two apart whose product is $440$.
Guess near $\sqrt{440}\approx 21$.
Try the pair $(20, 22)$.

$$20 \times 22 = 440 \;\checkmark$$

💡 $\sqrt{440}\approx 21$, so the two factors hug $21$ — try $20$ and $22$.

#6 Guess and Check 6.EE.B.7 Step 4

Match the pair to the equation.
$n+1 = 20$ gives $n = 19$, and $n+3 = 22$ also gives $n = 19$.
Consistent.

$$n + 1 = 20 \;\Rightarrow\; n = 19$$

💡 Both equations agree, so $n = 19$ is the right fit.

#3 Eliminate Possibilities 2.NBT.A.1 Step 5

Add the digits of $n = 19$.
Sum $= 1 + 9 = 10$.
Match against the choices: this is choice (C).

$$1 + 9 = 10 \;\Rightarrow\; \textbf{(C)}$$

💡 Digit sum of a two-digit number is just tens-place plus ones-place.

[1] #9 5.OA.A.2 Notice $(n+1)! = (n+1)\cdot n!$ and $(n+2)! = (n+2)(n+1)\cdot n!$. So both terms

[2] #9 6.EE.A.3 Divide both sides by $n!$ — it cancels cleanly. What is left is a product of two

[3] #6 4.OA.B.4 Look for two whole numbers two apart whose product is $440$. Guess near $\sqrt{4

[4] #6 6.EE.B.7 Match the pair to the equation. $n+1 = 20$ gives $n = 19$, and $n+3 = 22$ also g

[5] #3 2.NBT.A.1 Add the digits of $n = 19$. Sum $= 1 + 9 = 10$. Match against the choices: this

Review

Reasonableness: Quick check: $(19+1)!+(19+2)! = 20! + 21! = 20!\cdot(1+21) = 20!\cdot 22$, and $19!\cdot 440 = 19!\cdot 20\cdot 22 = 20!\cdot 22$. Both sides equal. Answer $n=19$ is verified and digit sum $10$ matches choice (C).

Alternative: Tool #13 (Algebra): after $(n+1)(n+3)=440$ expand to $n^2+4n+3 = 440$, so $n^2+4n-437 = 0$. The quadratic formula gives $n = \frac{-4+\sqrt{16+1748}}{2} = \frac{-4+42}{2} = 19$. Same answer, more work.

CCSS standards used (min grade 6)

5.OA.A.2 Write simple expressions that record calculations with numbers (Rewriting $(n+1)!$ and $(n+2)!$ in terms of $n!$ so the common factor pops out.)
6.EE.A.3 Apply the properties of operations to generate equivalent expressions (Factoring $(n+1)$ out of $(n+1)+(n+2)(n+1)$ to turn the equation into $(n+1)(n+3)=440$.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Hunting for two whole-number factors of $440$ that differ by $2$.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Solving the tiny linear equation $n+1=20$ for $n=19$.)
2.NBT.A.1 Understand that the three digits of a three-digit number represent hundreds, tens, and ones (Reading the digits of $19$ as tens-and-ones to add them.)

⭐ This AMC 10 problem only needs Grade 6 factoring you already know: pull the shared $n!$ out of both terms, get $(n+1)(n+3)=440$, spot $20\times 22$, so $n=19$ and the digit sum is $10$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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