AMC 10 · 2021 · #12
Grade 6 rate-ratioProblem
Let N=34⋅34⋅63⋅270. What is the ratio of the sum of the odd divisors of N to the sum of the even divisors of N?
Pick an answer.
AMC 10 2021 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: Let $N = 34 \cdot 34 \cdot 63 \cdot 270$. Find the ratio (sum of odd divisors of $N$) $:$ (sum of even divisors of $N$).
Givens: $N = 34 \cdot 34 \cdot 63 \cdot 270$; Answer choices: $1:16,\; 1:15,\; 1:14,\; 1:8,\; 1:3$
Unknowns: Ratio of sum of odd divisors to sum of even divisors of $N$
Understand
Restated: Let $N = 34 \cdot 34 \cdot 63 \cdot 270$. Find the ratio (sum of odd divisors of $N$) $:$ (sum of even divisors of $N$).
Givens: $N = 34 \cdot 34 \cdot 63 \cdot 270$; Answer choices: $1:16,\; 1:15,\; 1:14,\; 1:8,\; 1:3$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #9 Solve an Easier Related Problem, #5 Look for a Pattern, #3 Eliminate Possibilities
Tool #7 (Subproblems) splits the work into three pieces: (a) prime-factorize $N$ to find the power of $2$ in it; (b) sum the odd divisors using only the odd prime part; (c) get the sum of even divisors by subtracting from the total. Tool #9 (Easier Problem) checks the key insight on a tiny case like $N = 2^3 \cdot 3$ first, where you can list every divisor by hand. Tool #5 (Pattern) recognizes that every divisor is (power of $2$) $\times$ (odd divisor), so the even-divisor sum is $(2 + 4 + 8) = 14$ times the odd-divisor sum. Tool #3 (Eliminate) confirms (C) against the five ratio choices.
Execute — Answer: C
6.NS.B.4 Step 1 - Prime-factorize each factor of $N$.
- $34 = 2 \cdot 17$, so $34 \cdot 34 = 2^2 \cdot 17^2$.
- $63 = 9 \cdot 7 = 3^2 \cdot 7$.
- $270 = 27 \cdot 10 = 3^3 \cdot 2 \cdot 5$.
- Multiply and collect the same primes.
💡 Grade 6 prime factorization: break each factor into primes, then merge.
4.OA.B.4 Step 2 - Test the key idea on a tiny case.
- Let $M = 2^3 \cdot 3 = 24$.
- Divisors of $24$: $1, 2, 3, 4, 6, 8, 12, 24$.
- Odd divisors: $1, 3$, sum $= 4$.
- Even divisors: $2, 4, 6, 8, 12, 24$, sum $= 56$.
- Ratio $= 4 : 56 = 1 : 14$.
- Notice $14 = 2 + 4 + 8$: every even divisor is (some odd divisor) $\times$ ($2$ or $4$ or $8$).
💡 Grade 4 divisors: a small case verifies the structure before generalizing.
6.EE.A.3 Step 3 - Generalize.
- Let $S_{odd}$ be the sum of odd divisors of $N$.
- Every divisor of $N$ is (a power of $2$ from $2^0$ to $2^3$) $\times$ (an odd divisor).
- So the total sum of divisors is $\sigma(N) = (2^0 + 2^1 + 2^2 + 2^3) \cdot S_{odd} = 15 \, S_{odd}$.
- The even part drops the $2^0 = 1$ term: sum of even divisors $= (2 + 4 + 8) \cdot S_{odd} = 14 \, S_{odd}$.
💡 Grade 6 distributive thinking: factor (power of $2$) out of every divisor.
6.RP.A.1 Step 4 - Form the ratio.
- $S_{odd} : S_{even} = S_{odd} : 14 \, S_{odd} = 1 : 14$.
- The specific value of $S_{odd}$ (which depends on the $3, 5, 7, 17$ factors) cancels out of the ratio.
💡 Grade 6 ratios: the unknown $S_{odd}$ cancels, leaving a pure ratio.
6.RP.A.3 Step 5 - Match to the choices: $1 : 14$ is (C).
- The other options ($1:16, 1:15, 1:8, 1:3$) correspond to mistakes like including $2^0$ in the even-divisor sum or using a different power of $2$ in $N$.
💡 Grade 6 ratio matching: pick the listed ratio that equals $1 : 14$.
6.NS.B.4 Prime-factorize each factor of $N$. $34 = 2 \cdot 17$, so $34 \cdot 34 = 2^2 \cd 4.OA.B.4 Test the key idea on a tiny case. Let $M = 2^3 \cdot 3 = 24$. Divisors of $24$: 6.EE.A.3 Generalize. Let $S_{odd}$ be the sum of odd divisors of $N$. Every divisor of $N 6.RP.A.1 Form the ratio. $S_{odd} : S_{even} = S_{odd} : 14 \, S_{odd} = 1 : 14$. The spe 6.RP.A.3 Match to the choices: $1 : 14$ is (C). The other options ($1:16, 1:15, 1:8, 1:3$ Review
Reasonableness: Sanity check the power of $2$: $34 \cdot 34$ contributes $2^2$, and $270$ contributes one more $2$, so $N$ has exactly $2^3$. Total divisor sum formula $\sigma(2^3) = 1 + 2 + 4 + 8 = 15$ factors as $1 + 14$, matching odd-vs-even. The ratio $1 : 14$ is the same as for any number of the form $2^3 \cdot (\text{odd})$, which makes sense: only the power of $2$ in $N$ controls this ratio.
Alternative: Tool #13 (Algebra) with the divisor-sum formula. $\sigma(N) = \sigma(2^3) \cdot \sigma(3^5) \cdot \sigma(5) \cdot \sigma(7) \cdot \sigma(17^2)$. The odd-divisor sum is the same product without $\sigma(2^3)$, i.e. $S_{odd} = \sigma(N)/15$. Then $S_{even} = \sigma(N) - S_{odd} = \sigma(N)(1 - 1/15) = 14 \sigma(N)/15 = 14 \, S_{odd}$. Same ratio $1 : 14$, choice (C).
CCSS standards used (min grade 6)
4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Listing all divisors of the test case $M = 24$ to verify the structure.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Prime-factorizing $34, 63, 270$ and combining into $N = 2^3 \cdot 3^5 \cdot 5 \cdot 7 \cdot 17^2$.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Factoring (power of $2$) out of every divisor to write $S_{even} = (2+4+8) \, S_{odd}$.)6.RP.A.1Understand the concept of a ratio and use ratio language (Forming and simplifying the ratio $S_{odd} : S_{even}$.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Matching the simplified ratio $1 : 14$ to the answer choices.)
⭐ This AMC 10 problem only needs Grade 6 ratio sense you already know! Prime-factorize $N = 2^3 \cdot 3^5 \cdot 5 \cdot 7 \cdot 17^2$ — only the $2^3$ part matters. Every even divisor is $(2 \text{ or } 4 \text{ or } 8) \times (\text{odd divisor})$, so the even-divisor sum is $(2+4+8) = 14$ times the odd-divisor sum. Ratio $1 : 14$, answer (C).
⭐ This AMC 10 problem only needs Grade 6 ratio sense you already know! Prime-factorize $N = 2^3 \cdot 3^5 \cdot 5 \cdot 7 \cdot 17^2$ — only the $2^3$ part matters. Every even divisor is $(2 \text{ or } 4 \text{ or } 8) \times (\text{odd divisor})$, so the even-divisor sum is $(2+4+8) = 14$ times the odd-divisor sum. Ratio $1 : 14$, answer (C).
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