AMC 8 · 2022 · #17
Easy mode Grade 4Problem
There is a special symbol called the double factorial, written n!!. When n is an even positive integer, n!! means you multiply together all the even numbers from 2 up to n.
For example, 8!!=2⋅4⋅6⋅8.
Now picture a long sum. Start with 2!!, then add 4!!, then 6!!, and keep going by twos all the way up to 2022!!:
2!!+4!!+6!!+⋯+2018!!+2020!!+2022!!
This sum is a huge number. We do not need the whole number — we only want its very last digit (the units digit).
What is the units digit of this sum?
Pick an answer.
AMC 8 2022 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: 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
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) $8$
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
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.
💡 Multiplying a small multi-digit number like $48$ by a one-digit number like $8$ is exactly the Grade 4 multi-digit multiplication skill.
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 $0$.
💡 Spotting that multiplying by $10$ tacks a $0$ onto the end is the Grade 3 "identify arithmetic patterns" skill in action.
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.
💡 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.
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).
💡 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.
4.NBT.B.5 Compute the first few double factorials directly to see their units digits. Buil 3.OA.D.9 Read off the units digits and look for a pattern: $2,\ 8,\ 8,\ 4$. Now compute o 3.OA.D.9 Generalize: for any even $n \geq 10$, the product $n!! = 2 \cdot 4 \cdot 6 \cdot 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.5Multiply 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.9Identify 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.5Fluently 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!
⭐ This AMC 8 problem only needs Grade 4 multiplication and one pattern you already know — anything times 10 ends in 0!
More like this
Same archetype — closest grade level first.