AMC 8 · 2022 · #17

Grade 4 arithmeticnumber-theory

units-digit-trackingpattern-recognitionmultiples modular-arithmetic-mod-10pattern-recognition ↑ Prerequisites: units-digit-trackingmultiples

📏 Short solution 💡 2 insights

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Problem

If $n$ is an even positive integer, the $\emph{double factorial}$ notation $n!!$ represents the product of all the even integers from $2$ to $n$ . For example, $8!! = 2 \cdot 4 \cdot 6 \cdot 8$ . What is the units digit of the following sum? $2!! + 4!! + 6!! + \cdots + 2018!! + 2020!! + 2022!!$

$\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: For an even positive integer $n$, the double factorial $n!!$ means the product of all even integers from $2$ up to $n$ (so $8!! = 2 \cdot 4 \cdot 6 \cdot 8$). What is the units digit (the ones place) of the sum $2!! + 4!! + 6!! + \cdots + 2020!! + 2022!!$?

Givens: $n!!$ is defined for even $n$ as the product of all even integers from $2$ to $n$; The sum runs over every even $n$ from $2$ up to $2022$ (so $1011$ terms total); Answer choices: (A) $0$, (B) $2$, (C) $4$, (D) $6$, (E) $8$

Unknowns: The units digit (ones-place digit) of the long sum $2!! + 4!! + 6!! + \cdots + 2022!!$

Understand

Plan

Primary tool: #5 Look for a Pattern

Secondary: #9 Solve an Easier Related Problem

The sum has $1011$ terms and the last term $2022!!$ is astronomically large, so direct calculation is hopeless. Tool #9 (Easier Related Problem) says: compute just the first few double factorials and watch what happens to their units digits. Tool #5 (Look for a Pattern) then takes over: once we see that $10!!$ ends in $0$, the factor $10$ inside every later $n!!$ ($n \geq 10$) guarantees a units digit of $0$ for every later term, so only the first four terms ($2!!, 4!!, 6!!, 8!!$) can influence the answer. After that the problem collapses to a one-line addition.

Execute — Answer: B

#9 Solve an Easier Related Problem 4.NBT.B.5 Step 1

Compute the first few double factorials directly to see their units digits.
Building each from the previous by multiplying one more even number keeps the arithmetic small.

$$2!! = 2,\ 4!! = 2 \cdot 4 = 8,\ 6!! = 8 \cdot 6 = 48,\ 8!! = 48 \cdot 8 = 384$$

💡 Multiplying a small multi-digit number like $48$ by a one-digit number like $8$ is exactly the Grade 4 multi-digit multiplication skill.

#5 Look for a Pattern 3.OA.D.9 Step 2

Read off the units digits and look for a pattern: $2,\ 8,\ 8,\ 4$.
Now compute one more term to see what happens at $n = 10$: $10!! = 8!! \cdot 10 = 384 \cdot 10 = 3840$, which ends in $0$.

$$\text{Units digits so far: } 2,\; 8,\; 8,\; 4,\; 0$$

💡 Spotting that multiplying by $10$ tacks a $0$ onto the end is the Grade 3 "identify arithmetic patterns" skill in action.

#5 Look for a Pattern 3.OA.D.9 Step 3

Generalize: for any even $n \geq 10$, the product $n!! = 2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdots n$ contains the factor $10$.
Any whole number multiplied by $10$ has units digit $0$, so every term from $10!!$ all the way up to $2022!!$ contributes a $0$ in the ones place.

$$10!!,\ 12!!,\ 14!!,\ \ldots,\ 2022!! \;\Rightarrow\; \text{all end in } 0$$

💡 Recognizing the rule "any product with a factor of $10$ ends in $0$" is a Grade 3 multiplication-pattern observation, and it makes thousands of terms drop out at once.

#5 Look for a Pattern 2.NBT.B.5 Step 4

Because adding $0$ never changes a units digit, only $2!! + 4!! + 6!! + 8!!$ matters.
Add their units digits: $2 + 8 + 8 + 4 = 22$.
The units digit of $22$ is $2$, so the entire sum ends in $2$.
This matches choice (B).

$$2 + 8 + 8 + 4 = 22 \;\Rightarrow\; \text{units digit } = 2 \;\Rightarrow\; \textbf{(B)}$$

💡 Adding four single-digit numbers and reading the ones place is a Grade 2 fluency skill — the hard work was all in the pattern, not the arithmetic.

[1] #9 4.NBT.B.5 Compute the first few double factorials directly to see their units digits. Buil

[2] #5 3.OA.D.9 Read off the units digits and look for a pattern: $2,\ 8,\ 8,\ 4$. Now compute o

[3] #5 3.OA.D.9 Generalize: for any even $n \geq 10$, the product $n!! = 2 \cdot 4 \cdot 6 \cdot

[4] #5 2.NBT.B.5 Because adding $0$ never changes a units digit, only $2!! + 4!! + 6!! + 8!!$ mat

Review

Reasonableness: The units digit must be one of $\{0,1,2,\ldots,9\}$, and our answer $2$ is in that range. As a sanity check, compute the actual first four terms: $2 + 8 + 48 + 384 = 442$, whose units digit is $2$. Every later term ends in $0$, so adding any number of them to $442$ leaves the units digit at $2$. The answer is consistent.

Alternative: Tool #3 (Eliminate Possibilities) could verify: $2 + 8 + 48 + 384 = 442$ already ends in $2$, immediately killing choices (A) $0$, (C) $4$, (D) $6$, and (E) $8$. Tool #7 (Identify Subproblems) is another framing — split the sum into the "first four terms that matter" and the "all-zero-units tail" and handle each piece on its own.

CCSS standards used (min grade 4)

4.NBT.B.5 Multiply a whole number of up to four digits by a one-digit whole number (Computing the first few double factorials by hand: $48 \cdot 8 = 384$ (and $384 \cdot 10 = 3840$) — multi-digit times one-digit multiplication.)
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations (Noticing that once the factor $10$ appears in the product, every later $n!!$ has units digit $0$ — the key pattern that collapses $1011$ terms down to four.)
2.NBT.B.5 Fluently add and subtract within 100 (Adding the four relevant units digits $2 + 8 + 8 + 4 = 22$ to read off the final units digit.)

⭐ This AMC 8 problem only needs Grade 4 multiplication and one pattern you already know — anything times 10 ends in 0!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 4)

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