AMC 10 · 2019 · #11
Grade 6 arithmeticProblem
How many positive integer divisors of 2019 are perfect squares or perfect cubes (or both)?
Pick an answer.
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.
Toolkit + CCSS Solution
Understand
Restated: $201 = 3 \times 67$, so $201^9 = 3^9 \cdot 67^9$. Count the positive divisors of $201^9$ that are perfect squares, perfect cubes, or both.
Givens: $201^9 = 3^9 \cdot 67^9$; A divisor has the form $3^a \cdot 67^b$ with $0 \le a, b \le 9$; A perfect square needs every exponent even; A perfect cube needs every exponent a multiple of $3$; Answer choices: $32, 36, 37, 39, 41$
Unknowns: The number of divisors that are a perfect square or a perfect cube
Understand
Restated: $201 = 3 \times 67$, so $201^9 = 3^9 \cdot 67^9$. Count the positive divisors of $201^9$ that are perfect squares, perfect cubes, or both.
Givens: $201^9 = 3^9 \cdot 67^9$; A divisor has the form $3^a \cdot 67^b$ with $0 \le a, b \le 9$; A perfect square needs every exponent even; A perfect cube needs every exponent a multiple of $3$; Answer choices: $32, 36, 37, 39, 41$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #12 Draw a Venn Diagram, #16 Change Focus / Count the Complement, #3 Eliminate Possibilities
Tool #2 (Systematic List): list the legal exponents $a, b$ for each shape — even values for squares, multiples of $3$ for cubes. Tool #12 (Venn): squares and cubes overlap on $6$th powers, so use inclusion-exclusion: $|\text{sq} \cup \text{cu}| = |\text{sq}| + |\text{cu}| - |\text{both}|$. Tool #16 (Complement) frames "both" as the overlap to subtract. Tool #3 matches the final count to the five choices.
Execute — Answer: C
6.EE.A.1 Step 1 - Factor $201$ to see the shape of every divisor.
- $201 = 3 \times 67$ with $67$ prime, so $201^9 = 3^9 \cdot 67^9$.
- Every positive divisor has the form $3^a \cdot 67^b$ with $0 \le a \le 9$ and $0 \le b \le 9$.
💡 Grade 6 exponent expressions: knowing the prime factor shape turns the count into picking $a$ and $b$.
6.EE.A.1 Step 2 - List the exponents that make $3^a \cdot 67^b$ a perfect square: every prime power must be even.
- The even values of $a$ in $\{0, \dots, 9\}$ are $\{0, 2, 4, 6, 8\}$ — five choices.
- Same for $b$.
- Multiply: $5 \times 5 = 25$ perfect-square divisors.
💡 Grade 6: even exponents on each prime give a perfect square, so just count even values in $0$ to $9$.
6.EE.A.1 Step 3 - List the exponents that make $3^a \cdot 67^b$ a perfect cube: every prime power must be a multiple of $3$.
- The multiples of $3$ in $\{0, \dots, 9\}$ are $\{0, 3, 6, 9\}$ — four choices.
- Multiply: $4 \times 4 = 16$ perfect-cube divisors.
💡 Grade 6: multiples of $3$ on each prime give a perfect cube, so count multiples of $3$ in $0$ to $9$.
6.NS.B.4 Step 4 - Count the overlap.
- A divisor that is both a square and a cube needs each exponent to be a multiple of $6$ (an even multiple of $3$).
- The multiples of $6$ in $\{0, \dots, 9\}$ are $\{0, 6\}$ — two choices.
- Multiply: $2 \times 2 = 4$ divisors that are sixth powers (square and cube at once).
💡 Grade 6 LCM: the smallest common requirement of even and multiple-of-$3$ is multiple-of-$6$.
4.OA.A.3 Step 5 Apply inclusion-exclusion (the Venn count): the union is squares plus cubes minus the overlap, so $25 + 16 - 4 = 37$.
💡 Grade 4 multi-step word problem: add the two groups, take away the part double-counted.
4.NBT.A.2 Step 6 Match $37$ to the answer choices: $(C)$.
💡 Grade 4: read the list and pick the matching number.
6.EE.A.1 Factor $201$ to see the shape of every divisor. $201 = 3 \times 67$ with $67$ pr 6.EE.A.1 List the exponents that make $3^a \cdot 67^b$ a perfect square: every prime powe 6.EE.A.1 List the exponents that make $3^a \cdot 67^b$ a perfect cube: every prime power 6.NS.B.4 Count the overlap. A divisor that is both a square and a cube needs each exponen 4.OA.A.3 Apply inclusion-exclusion (the Venn count): the union is squares plus cubes minu 4.NBT.A.2 Match $37$ to the answer choices: $(C)$. Review
Reasonableness: Squares (25) and cubes (16) are both small subsets of the $10 \times 10 = 100$ total divisors, and they overlap on a tiny set (the $4$ sixth powers). The union should sit between $25$ and $25 + 16 = 41$, and closer to $41$ since the overlap is small. $37$ lands right in that band.
Alternative: Tool #6 (Guess and Check) by direct enumeration: list every $(a, b)$ pair with $0 \le a, b \le 9$ and mark which are squares, cubes, or both. With $100$ pairs this is tedious but mechanical, and confirms the same $37$ shaded cells.
CCSS standards used (min grade 6)
4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Matching the final count $37$ to the answer choices.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Combining the three counts with $25 + 16 - 4 = 37$.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Reading divisors as $3^a \cdot 67^b$ and counting even and multiple-of-$3$ exponents.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Seeing that being a square and a cube at the same time means each exponent is a multiple of $6 = \mathrm{lcm}(2, 3)$.)
⭐ This AMC 10 problem only needs Grade 6 exponents and LCM you already know! Write $201^9 = 3^9 \cdot 67^9$, so each divisor is $3^a \cdot 67^b$. Squares need even $a, b$ ($5 \times 5 = 25$); cubes need $a, b$ multiples of $3$ ($4 \times 4 = 16$); both at once needs multiples of $6$ ($2 \times 2 = 4$). Add and subtract the overlap: $25 + 16 - 4 = \mathbf{37}$, answer $(C)$.
⭐ This AMC 10 problem only needs Grade 6 exponents and LCM you already know! Write $201^9 = 3^9 \cdot 67^9$, so each divisor is $3^a \cdot 67^b$. Squares need even $a, b$ ($5 \times 5 = 25$); cubes need $a, b$ multiples of $3$ ($4 \times 4 = 16$); both at once needs multiples of $6$ ($2 \times 2 = 4$). Add and subtract the overlap: $25 + 16 - 4 = \mathbf{37}$, answer $(C)$.
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