AMC 8 · 2017 · #19

Grade 6 number-theory

Archetype Arithmetic Expression Decomposition

factorialprime-factorizationexponentsfactors identify-subproblemspattern-recognition ↑ Prerequisites: factorialprime-factorizationexponents

📏 Medium solution 💡 4 insights

Problem

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 $5^n$ is a factor of the sum $98!+99!+100!$ ?

$\textbf{(A) }23 \qquad \textbf{(B) }24 \qquad \textbf{(C) }25 \qquad \textbf{(D) }26 \qquad \textbf{(E) }27$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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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

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

#7 Identify Subproblems 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.

$$98! + 99! + 100! = 98!\,(1 + 99 + 100 \cdot 99)$$

💡 Pulling out the common factor $98!$ is the distributive property — turning a sum into a product so we can count primes in each piece.

#7 Identify Subproblems 4.NBT.B.5 Step 2

Simplify the bracket.
$100 \cdot 99 = 9900$, then $1 + 99 + 9900 = 10000$.
So the sum collapses to $98! \times 10000$.

$$1 + 99 + 100 \cdot 99 = 1 + 99 + 9900 = 10000$$

💡 Multi-digit multiplication and addition are Grade 4 skills — the only "trick" is choosing to factor first.

#7 Identify Subproblems 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.

$$10000 = 10^4 = (2 \cdot 5)^4 = 2^4 \cdot 5^4$$

💡 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.

#5 Look for a Pattern 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.

$\text{multiples of } 5\text{ up to }98: \lfloor 98/5 \rfloor = 19$;\; $\text{multiples of } 25: \lfloor 98/25 \rfloor = 3$;\; total $= 19 + 3 = 22$

💡 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.

#3 Eliminate Possibilities 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.

$$n = 22 + 4 = 26 \;\Rightarrow\; \textbf{(D)}$$

💡 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.

[1] #7 6.EE.A.3 Factor out the smallest factorial, $98!$, from every term. Since $99! = 99 \cdot

[2] #7 4.NBT.B.5 Simplify the bracket. $100 \cdot 99 = 9900$, then $1 + 99 + 9900 = 10000$. So th

[3] #7 6.EE.A.1 Count the $5$s inside $10000$. Write $10000$ as a power of ten, then split each

[4] #5 6.NS.B.4 Count the $5$s inside $98!$ by spotting the pattern. Among $1, 2, \ldots, 98$, e

[5] #3 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.5 Multiply 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.1 Write 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.3 Apply the properties of operations to generate equivalent expressions (Factoring the common $98!$ out of $98! + 99! + 100!$ using the distributive property.)
6.NS.B.4 Find 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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