AMC 10 · 2006 · #11
Grade 5 arithmeticWhat is the tens digit in the sum 7!+8!+9!+...+2006!
Pick an answer.
AMC 10 2006 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: Add up every factorial from $7!$ through $2006!$, that is $7!+8!+9!+\cdots+2006!$. What is the tens digit of this enormous total?
Givens: The sum runs over every whole number from $7$ to $2006$: $7!+8!+9!+\cdots+2006!$; A factorial $n!$ means the product $1\times 2\times 3\times\cdots\times n$; Only the tens digit of the final sum is wanted, not the whole number; Answer choices: (A) $1$, (B) $3$, (C) $4$, (D) $6$, (E) $9$
Unknowns: The tens digit (the second digit from the right) of the total sum
Understand
Restated: Add up every factorial from $7!$ through $2006!$, that is $7!+8!+9!+\cdots+2006!$. What is the tens digit of this enormous total?
Givens: The sum runs over every whole number from $7$ to $2006$: $7!+8!+9!+\cdots+2006!$; A factorial $n!$ means the product $1\times 2\times 3\times\cdots\times n$; Only the tens digit of the final sum is wanted, not the whole number; Answer choices: (A) $1$, (B) $3$, (C) $4$, (D) $6$, (E) $9$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #5 Look for a Pattern, #16 Change Focus / Count the Complement
The sum stretches to $2006!$, a number with thousands of digits, so Tool #9 (Solve an Easier Related Problem) is the key: because only the tens digit is asked, almost every term can be thrown away, shrinking the problem to a tiny sum. Tool #5 (Look for a Pattern) supplies the reason it shrinks — factorials pick up more and more trailing zeros as they grow, so from $10!$ onward each term ends in $00$ and cannot touch the tens digit. Tool #16 (Change Focus) keeps attention on just the last two digits at every step, so no giant multiplications or additions are ever needed.
Execute — Answer: C
4.OA.B.4 Step 1 See why $10!$ ends in two zeros
- A number ends in two zeros exactly when it is a multiple of $100=4\times 25$.
- Look at $10!=1\times2\times3\times\cdots\times10$.
- Among its factors are $5$ and $10$, which together supply the two fives needed for $25$, and there are more than enough even factors ($2,4,6,8$) to supply the $4$.
- So $10!$ is a multiple of $100$; in fact $10!=3628800$, which does end in $00$.
💡 Once a product collects a $10$ and another factor of $5$ paired with even numbers, it must end in two zeros.
5.NBT.A.2 Step 2 Every larger factorial keeps those zeros
- Each factorial past $10!$ is just $10!$ multiplied by more whole numbers: $11!=11\times10!$, $12!=12\times11\times10!$, and so on up to $2006!$.
- Multiplying a number that already ends in $00$ by any whole number leaves it ending in $00$.
- So every one of $10!,11!,12!,\ldots,2006!$ ends in $00$, and none of them changes the tens or units digit of the sum.
💡 Trailing zeros never disappear when you multiply by whole numbers, so once a term ends in $00$ it stays harmless forever.
5.NBT.B.5 Step 3 Keep only the three terms that matter
- Since every term from $10!$ on ends in $00$, the tens digit of the whole sum is decided by the only terms smaller than $10!$: namely $7!$, $8!$, and $9!$.
- Compute them: $7!=5040$, $8!=8\times5040=40320$, and $9!=9\times40320=362880$.
💡 Throwing away every term that ends in $00$ collapses a $2000$-term sum down to just three numbers.
5.NBT.A.1 Step 4 Add just the last two digits
- To find the tens digit, keep only the last two digits of each surviving term: $7!$ ends in $40$, $8!$ ends in $20$, and $9!$ ends in $80$.
- Add these: $40+20+80=140$, which itself ends in $40$.
- So the whole sum $7!+8!+\cdots+2006!$ ends in $40$.
- The tens digit is the $4$ in $40$, so the answer is (C).
💡 The last two digits of a sum depend only on the last two digits of its parts, so a two-digit addition settles everything.
4.OA.B.4 A number ends in two zeros exactly when it is a multiple of $100=4\times 25$. Lo 5.NBT.A.2 Each factorial past $10!$ is just $10!$ multiplied by more whole numbers: $11!=1 5.NBT.B.5 Since every term from $10!$ on ends in $00$, the tens digit of the whole sum is 5.NBT.A.1 To find the tens digit, keep only the last two digits of each surviving term: $7 Review
Reasonableness: Check the discarded terms really are safe: $10!=3628800$ ends in $00$, and every factorial after it is a multiple of $10!$, so it also ends in $00$ — confirmed, they cannot affect the tens digit. Check the surviving arithmetic against the full numbers: $5040+40320+362880=408240$, which indeed ends in $40$ with tens digit $4$, matching the shortcut. The answer $4$ is choice (C), one of the listed options, so it is consistent.
Alternative: Instead of tracking trailing zeros, work in modular arithmetic and reduce the whole sum modulo $100$ from the start. Every term with $n\ge 10$ satisfies $n!\equiv 0\pmod{100}$, so $7!+8!+\cdots+2006!\equiv 7!+8!+9!\pmod{100}\equiv 40+20+80\equiv 140\equiv 40\pmod{100}$. The residue $40$ gives tens digit $4$, choice (C).
CCSS standards used (min grade 5)
4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Recognizing that $10!$ contains the factors ($5$, $10$, and even numbers) that make it a multiple of $100$, so it ends in two zeros.)5.NBT.A.2Explain patterns in the number of zeros when multiplying by whole numbers (Arguing that every factorial past $10!$ is $10!$ times more whole numbers, so it keeps its two trailing zeros and stays $\equiv 00\pmod{100}$.)5.NBT.B.5Fluently multiply multi-digit whole numbers (Computing the three surviving factorials $7!=5040$, $8!=40320$, and $9!=362880$.)5.NBT.A.1Understand place value: identify the tens digit of a number (Adding the last two digits $40+20+80=140$ and reading the tens digit $4$ from the result $40$.)
⭐ When only the last digits matter, drop every number that ends in zeros — here all factorials from $10!$ up vanish, leaving just $7!+8!+9!$, which ends in $40$, so the tens digit is $4$.
⭐ When only the last digits matter, drop every number that ends in zeros — here all factorials from $10!$ up vanish, leaving just $7!+8!+9!$, which ends in $40$, so the tens digit is $4$.
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