AMC 8 · 2022 · #3
Easy mode Grade 4Problem
Imagine picking three whole numbers and multiplying them together. We will call the numbers , , and .
The three numbers must all be different. Write them in order from smallest to largest, so . When you multiply them, you should get . That is, .
How many different sets of three numbers will work?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: Count the number of triples of positive integers $(a, b, c)$ with $a < b < c$ whose product $a \cdot b \cdot c$ equals $100$.
Givens: $a$, $b$, $c$ are positive integers; $a \cdot b \cdot c = 100$; $a < b < c$ (strictly increasing); Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Unknowns: The number of triples $(a, b, c)$ that satisfy all of the conditions
Understand
Restated: Count the number of triples of positive integers $(a, b, c)$ with $a < b < c$ whose product $a \cdot b \cdot c$ equals $100$.
Givens: $a$, $b$, $c$ are positive integers; $a \cdot b \cdot c = 100$; $a < b < c$ (strictly increasing); Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #7 Identify Subproblems
"How many ways" with a small finite product ($100$) is a textbook trigger for Tool #2 (Systematic List). To keep the listing organized and guarantee we miss nothing, we use Tool #7 (Identify Subproblems) to split the big question into cases by the smallest value $a$. Because $a < b < c$ forces $a^3 < a \cdot b \cdot c = 100$, only $a \in \{1, 2, 3, 4\}$ are even possible, and within those only divisors of $100$ work — so the casework is tiny ($a = 1, 2, 4$), and inside each case the remaining list of factor pairs of $100/a$ is short enough to write out completely.
Execute — Answer: E
4.OA.B.4 Step 1 - Bound the smallest value $a$.
- Since $a < b$ and $a < c$, multiplying the three inequalities gives $a \cdot a \cdot a < a \cdot b \cdot c = 100$, so $a^3 < 100$.
- Checking cubes: $1^3 = 1$, $2^3 = 8$, $3^3 = 27$, $4^3 = 64$, $5^3 = 125$.
- So $a$ can only be $1, 2, 3,$ or $4$.
💡 Finding the small set of possible smallest factors is a Grade 4 factor-and-multiple move that shrinks the search space.
4.OA.B.4 Step 2 - Restrict to divisors of $100$.
- The divisors of $100 = 2^2 \cdot 5^2$ are $1, 2, 4, 5, 10, 20, 25, 50, 100$.
- Intersecting with $\{1, 2, 3, 4\}$ gives the only candidates $a \in \{1, 2, 4\}$ (note: $3$ is not a divisor of $100$, so it's dropped).
💡 Grade 4 students already find all factor pairs of a number, so spotting that $3$ isn't a factor of $100$ is enough to kill that case.
4.OA.B.4 Step 3 - Case $a = 1$.
- We need $b \cdot c = 100$ with $1 < b < c$.
- List factor pairs of $100$ in increasing order of $b$: $(2, 50), (4, 25), (5, 20), (10, 10)$.
- The first three satisfy $b < c$; the last has $b = c$ and is rejected.
💡 Listing factor pairs of $100$ in order is exactly the Grade 4 "find all factor pairs" skill.
4.OA.B.4 Step 4 - Case $a = 2$.
- We need $b \cdot c = 50$ with $2 < b < c$.
- Factor pairs of $50$ with $b \le c$: $(1, 50), (2, 25), (5, 10)$.
- Only $(5, 10)$ satisfies both $b > 2$ and $b < c$.
💡 Listing factor pairs of $50$ and filtering by $b > 2$ is the same Grade 4 factor-pair skill, just with one extra comparison.
4.OA.B.4 Step 5 - Case $a = 4$.
- We need $b \cdot c = 25$ with $4 < b < c$.
- The only factor pair of $25$ with $b \le c$ is $(5, 5)$, which fails $b < c$.
- So this case contributes nothing.
💡 Knowing $25 = 1 \times 25 = 5 \times 5$ as a Grade 4 factor pair makes it instant: the only candidate has $b = c$, which isn't allowed.
3.OA.D.8 Step 6 - Total the cases.
- Add the counts: $3 + 1 + 0 = 4$ triples.
- The four triples are $(1, 2, 50), (1, 4, 25), (1, 5, 20), (2, 5, 10)$.
- The answer is $4$, which is choice (E).
💡 Adding the case counts is a Grade 3 multi-step word-problem move — first multiply/factor inside each case, then add.
4.OA.B.4 Bound the smallest value $a$. Since $a < b$ and $a < c$, multiplying the three i 4.OA.B.4 Restrict to divisors of $100$. The divisors of $100 = 2^2 \cdot 5^2$ are $1, 2, 4.OA.B.4 Case $a = 1$. We need $b \cdot c = 100$ with $1 < b < c$. List factor pairs of $ 4.OA.B.4 Case $a = 2$. We need $b \cdot c = 50$ with $2 < b < c$. Factor pairs of $50$ wi 4.OA.B.4 Case $a = 4$. We need $b \cdot c = 25$ with $4 < b < c$. The only factor pair of 3.OA.D.8 Total the cases. Add the counts: $3 + 1 + 0 = 4$ triples. The four triples are $ Review
Reasonableness: Sanity check each triple by multiplying: $1 \cdot 2 \cdot 50 = 100$, $1 \cdot 4 \cdot 25 = 100$, $1 \cdot 5 \cdot 20 = 100$, $2 \cdot 5 \cdot 10 = 100$ — all four check out, and each has strictly increasing values. The bound $a^3 < 100$ also guarantees we couldn't have missed an $a \geq 5$ triple, so $4$ is the complete count and answer (E) fits.
Alternative: Tool #5 (Look for a Pattern) on prime factorizations: $100 = 2^2 \cdot 5^2$, so any triple is a way to split the four prime factors $\{2, 2, 5, 5\}$ into three groups (allowing empty groups, which contribute a factor of $1$). Enumerating splits up to ordering still gives the same four triples, but the systematic-list version above is cleaner for an AMC 8 student.
CCSS standards used (min grade 4)
4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Bounding $a \in \{1, 2, 4\}$ as factors of $100$ and listing the factor pairs of $100$, $50$, and $25$ in each case.)3.OA.D.8Solve two-step word problems using four operations within 100 (Adding the per-case counts $3 + 1 + 0 = 4$ at the end and verifying each candidate triple by multiplication.)
⭐ This AMC 8 problem only needs Grade 4 factor pairs you already know — just split $100$ into three growing pieces!
⭐ This AMC 8 problem only needs Grade 4 factor pairs you already know — just split $100$ into three growing pieces!