AMC 8 · 2017 · #19
Grade 6 number-theoryProblem
For any positive integer M, the notation M! denotes the product of the integers 1 through
M. What is the largest integer n for which 5n is a factor of the sum 98!+99!+100! ?
Pick an answer.
AMC 8 2017 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: Recall that $M! = 1 \cdot 2 \cdot 3 \cdots M$. We are told to find the largest integer $n$ such that $5^n$ divides the sum $98! + 99! + 100!$. Equivalently: how many copies of the prime $5$ appear in the prime factorization of that sum?
Givens: The expression is $98! + 99! + 100!$; $M!$ means the product of the integers from $1$ to $M$; We are looking for the highest power of $5$ that divides the whole sum; Answer choices: (A) $23$, (B) $24$, (C) $25$, (D) $26$, (E) $27$
Unknowns: The largest integer $n$ such that $5^n \mid (98! + 99! + 100!)$
Understand
Restated: Recall that $M! = 1 \cdot 2 \cdot 3 \cdots M$. We are told to find the largest integer $n$ such that $5^n$ divides the sum $98! + 99! + 100!$. Equivalently: how many copies of the prime $5$ appear in the prime factorization of that sum?
Givens: The expression is $98! + 99! + 100!$; $M!$ means the product of the integers from $1$ to $M$; We are looking for the highest power of $5$ that divides the whole sum; Answer choices: (A) $23$, (B) $24$, (C) $25$, (D) $26$, (E) $27$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #5 Look for a Pattern, #3 Eliminate Possibilities
Tool #7 (Identify Subproblems) is the workhorse here: the sum $98! + 99! + 100!$ shares a common factor $98!$, so we factor it out and split the problem into two independent counting jobs — "how many $5$s are inside $98!$?" and "how many $5$s are inside what's left in the parentheses?". Tool #5 (Look for a Pattern) handles the first subproblem: every $5$th number contributes one $5$, every $25$th number contributes an extra $5$, and so on — a clean, repeating pattern we can list. Tool #3 (Eliminate Possibilities) gives a final sanity check against the five answer choices.
Execute — Answer: D
6.EE.A.3 Step 1 - Factor out the smallest factorial, $98!$, from every term.
- Since $99! = 99 \cdot 98!$ and $100! = 100 \cdot 99 \cdot 98!$, the whole sum becomes $98!$ times a small whole number we can compute by hand.
💡 Pulling out the common factor $98!$ is the distributive property — turning a sum into a product so we can count primes in each piece.
4.NBT.B.5 Step 2 - Simplify the bracket.
- $100 \cdot 99 = 9900$, then $1 + 99 + 9900 = 10000$.
- So the sum collapses to $98! \times 10000$.
💡 Multi-digit multiplication and addition are Grade 4 skills — the only "trick" is choosing to factor first.
6.EE.A.1 Step 3 - Count the $5$s inside $10000$.
- Write $10000$ as a power of ten, then split each $10$ into $2 \times 5$.
- There are exactly four $5$s.
💡 Writing $10000$ as $10^4$ and expanding with exponent rules is the Grade 6 "whole-number exponents" idea — much faster than dividing by $5$ repeatedly.
6.NS.B.4 Step 4 - Count the $5$s inside $98!$ by spotting the pattern.
- Among $1, 2, \ldots, 98$, every $5$th number is a multiple of $5$, every $25$th number is a multiple of $25$ (and contributes a second $5$), and $125 > 98$ so we stop.
- Count the multiples and add.
💡 Each multiple of $5$ contributes one $5$; multiples of $25$ are double-counted because they hide an extra $5$ — the same prime-factor logic used for GCF and LCM.
6.EE.A.1 Step 5 - Add the two counts.
- Since $98! \times 10000$ is a product, its total number of $5$s is the sum of the $5$s in each factor: $22 + 4 = 26$.
- So $n = 26$, which is choice (D).
- Tool #3: choices (A)$23$, (B)$24$, (C)$25$, (E)$27$ are all eliminated — they would either ignore the extra $5$s from $10000$ or miss the $25$-multiples bonus.
💡 Exponents add when you multiply: $5^{22} \cdot 5^4 = 5^{26}$ — exactly the Grade 6 exponent rule applied to two numbers we already factored.
6.EE.A.3 Factor out the smallest factorial, $98!$, from every term. Since $99! = 99 \cdot 4.NBT.B.5 Simplify the bracket. $100 \cdot 99 = 9900$, then $1 + 99 + 9900 = 10000$. So th 6.EE.A.1 Count the $5$s inside $10000$. Write $10000$ as a power of ten, then split each 6.NS.B.4 Count the $5$s inside $98!$ by spotting the pattern. Among $1, 2, \ldots, 98$, e 6.EE.A.1 Add the two counts. Since $98! \times 10000$ is a product, its total number of $ Review
Reasonableness: The answer $26$ has to sit between the natural lower and upper bounds. Lower bound: $98!$ alone already has $22$ factors of $5$, so the sum (which is a multiple of $98!$) has at least $22$ — eliminating any answer below $22$. Upper bound: the bracketed factor $10000 = 2^4 \cdot 5^4$ adds exactly four more $5$s and nothing else introduces new $5$s, so the count tops out at $22 + 4 = 26$. The answer $26$ sits right at that bound, matching choice (D).
Alternative: Tool #9 (Easier Related Problem): try the same question on $3! + 4! + 5!$ to see the mechanism. Factor: $3!(1 + 4 + 4 \cdot 5) = 6 \cdot 25 = 150 = 2 \cdot 3 \cdot 5^2$. So the answer there is $2$. The bracket gives $1 + 4 + 20 = 25 = 5^2$, contributing both $5$s; $3!$ contributes none. This tiny version mirrors exactly what happens at $98! + 99! + 100!$: the bracket is what supplies the bonus $5$s on top of whatever the small factorial already had.
CCSS standards used (min grade 6)
4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Computing $100 \cdot 99 = 9900$ and $1 + 99 + 9900 = 10000$ to collapse the bracketed expression.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Rewriting $10000$ as $10^4 = 2^4 \cdot 5^4$ and combining $5^{22} \cdot 5^4 = 5^{26}$ via the exponent-addition rule.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Factoring the common $98!$ out of $98! + 99! + 100!$ using the distributive property.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Counting factors of the prime $5$ inside $98!$ by listing multiples of $5$ and of $25$ — the same prime-factor counting technique used for GCF/LCM.)
⭐ This AMC 8 problem only needs the Grade 6 exponent and factoring ideas you already know — pull out the common factor, then count the $5$s in each piece!
⭐ This AMC 8 problem only needs the Grade 6 exponent and factoring ideas you already know — pull out the common factor, then count the $5$s in each piece!
More like this
Same archetype — closest grade level first.