AMC 8 · 2022 · #3
Easy mode Grade 4Problem
Imagine picking three whole numbers and multiplying them together. We will call the numbers a, b, and c.
The three numbers must all be different. Write them in order from smallest to largest, so a<b<c. When you multiply them, you should get 100. That is, a×b×c=100.
How many different sets of three numbers will work?
Pick an answer.
AMC 8 2022 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: 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!
More like this
Same archetype — closest grade level first.