AMC 10 · 2019 · #2

Grade 5 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

factorialprime-factorizationlegendre-formuladigit-decompositionplace-value identify-subproblemspattern-recognition ↑ Prerequisites: factorialprime-factorization

📏 Short solution 💡 2 insights

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Problem

What is the hundreds digit of $(20!-15!)?$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2019 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.

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Toolkit + CCSS Solution

Understand

Restated: Find the hundreds digit of the number $20! - 15!$, where $n!$ means $1 \cdot 2 \cdot 3 \cdots n$.

Givens: $20! = 1 \cdot 2 \cdot 3 \cdots 20$; $15! = 1 \cdot 2 \cdot 3 \cdots 15$; We want only the hundreds digit (the digit in the $100$s place) of $20! - 15!$; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $4$, (E) $5$

Unknowns: The hundreds digit of $20! - 15!$

Understand

Restated: Find the hundreds digit of the number $20! - 15!$, where $n!$ means $1 \cdot 2 \cdot 3 \cdots n$.

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #5 Look for a Pattern, #16 Change Focus / Count the Complement

Tool #9 (Easier Problem): we cannot compute $20!$ or $15!$ directly, so we ask a smaller question — what are the last three digits of each? A number's hundreds digit is decided by its remainder when divided by $1000$. Tool #5 (Pattern): count factors of $10$ inside each factorial. Each pair of a $2$ and a $5$ inside the product gives one trailing zero. Tool #16 (Change Focus): instead of asking 'what is the hundreds digit of $20! - 15!$', we ask 'is $20! - 15!$ divisible by $1000$?' — if yes, the hundreds digit must be $0$.

Execute — Answer: A

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

Count the factors of $5$ inside $15!$.
The numbers $5$, $10$, and $15$ each contribute one factor of $5$, giving $3$ factors of $5$ total.
The number of factors of $2$ is much larger (every even number gives at least one), so we have at least $3$ matching pairs of $2$ and $5$.
Therefore $15!$ is divisible by $10^3 = 1000$.

$15! \supseteq 5 \cdot 10 \cdot 15 \;\Rightarrow\; 5^3 \mid 15!$, and $2^3 \mid 15!$ trivially $\Rightarrow\; 1000 \mid 15!$

💡 Each factor of $5$ paired with a factor of $2$ creates a trailing zero — $15!$ has at least three.

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

Count the factors of $5$ inside $20!$ the same way.
The numbers $5$, $10$, $15$, $20$ each give one factor of $5$, totaling $4$ factors of $5$.
So $20!$ is divisible by $10^4 = 10000$, and certainly by $1000$.

$20! \supseteq 5 \cdot 10 \cdot 15 \cdot 20 \;\Rightarrow\; 5^4 \mid 20!$, and $2^4 \mid 20!$ trivially $\Rightarrow\; 1000 \mid 20!$

💡 Same pattern as $15!$, just more factors of $5$ — $20!$ has even more trailing zeros.

#16 Change Focus / Count the Complement 5.NBT.A.1 Step 3

Both $20!$ and $15!$ are multiples of $1000$, so each ends in $000$.
The difference of two multiples of $1000$ is also a multiple of $1000$, so $20! - 15!$ ends in $000$.
The hundreds digit (the digit in the $100$s place of $\ldots 000$) is $0$.

$1000 \mid 20!,\ 1000 \mid 15! \;\Rightarrow\; 1000 \mid (20! - 15!)$, so the last three digits are $000 \;\Rightarrow\; \textbf{(A)}\ 0$

💡 If a number ends in $\ldots 000$, the hundreds place is $0$.

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

Sanity-check with a smaller version.
Take $4! - 2! = 24 - 2 = 22$.
Neither has a factor of $10$, so the units digit is not forced.
The principle is: when both factorials share enough factors of $10$, the trailing zeros survive the subtraction.
$15!$ already has three trailing zeros — that's exactly what we need.

Smaller test: $4! - 2! = 22$ (no shared trailing zeros). For $20! - 15!$, both share $\ge 3$ trailing zeros $\Rightarrow$ last three digits are $000$.

💡 Working with tiny factorials first builds confidence in the trailing-zero rule.

[1] #9 4.OA.B.4 Count the factors of $5$ inside $15!$. The numbers $5$, $10$, and $15$ each cont

[2] #5 4.OA.B.4 Count the factors of $5$ inside $20!$ the same way. The numbers $5$, $10$, $15$,

[3] #16 5.NBT.A.1 Both $20!$ and $15!$ are multiples of $1000$, so each ends in $000$. The differe

[4] #9 5.NBT.A.1 Sanity-check with a smaller version. Take $4! - 2! = 24 - 2 = 22$. Neither has a

Review

Reasonableness: Any number that is a multiple of $1000$ must end in three zeros, so its hundreds digit, tens digit, and units digit are each $0$. Since $20! - 15!$ is a multiple of $1000$, its hundreds digit is forced to be $0$. The other choices (B) $1$, (C) $2$, (D) $4$, (E) $5$ would each require the number to leave a non-zero remainder modulo $1000$, but it doesn't.

Alternative: Tool #13 (Algebra): write $20! - 15! = 15!\,(16 \cdot 17 \cdot 18 \cdot 19 \cdot 20 - 1)$. Since $15!$ is divisible by $1000$, the whole product is too. Same conclusion via factoring rather than digit-counting.

CCSS standards used (min grade 5)

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Counting the factors of $5$ (and $2$) inside $15!$ and $20!$ to confirm both are multiples of $1000$.)
5.NBT.A.1 Recognize that a digit represents ten times what it represents in place to its right (Reading the hundreds digit of a number that ends in $\ldots 000$ as $0$.)

⭐ This AMC 10 problem only needs Grade 5 "place value" you already know — both $20!$ and $15!$ end in at least three zeros, so their difference ends in three zeros and the hundreds digit is $0$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 5)

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