AMC 10 · 2002 · #1
Grade 8 rate-ratioThe ratio 6200222001⋅32003 is:
Pick an answer.
AMC 10 2002 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: Simplify the single fraction $\dfrac{2^{2001}\cdot 3^{2003}}{6^{2002}}$ to one of the given values.
Givens: The expression $\dfrac{2^{2001}\cdot 3^{2003}}{6^{2002}}$; The numerator is a power of $2$ times a power of $3$; the denominator is a power of $6$; Answer choices: (A) $\tfrac16$, (B) $\tfrac13$, (C) $\tfrac12$, (D) $\tfrac23$, (E) $\tfrac32$
Unknowns: The single value the fraction equals
Understand
Restated: Simplify the single fraction $\dfrac{2^{2001}\cdot 3^{2003}}{6^{2002}}$ to one of the given values.
Givens: The expression $\dfrac{2^{2001}\cdot 3^{2003}}{6^{2002}}$; The numerator is a power of $2$ times a power of $3$; the denominator is a power of $6$; Answer choices: (A) $\tfrac16$, (B) $\tfrac13$, (C) $\tfrac12$, (D) $\tfrac23$, (E) $\tfrac32$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #9 Solve an Easier Related Problem, #3 Eliminate Possibilities
The one obstacle is that the denominator's base $6$ does not match the numerator's bases $2$ and $3$. Tool #7 (Identify Subproblems) fixes this by splitting $6^{2002}$ into a base-$2$ piece and a base-$3$ piece, turning one scary fraction into two clean same-base quotients. Tool #9 (Solve an Easier Related Problem) then makes the giant exponents harmless: because the powers differ by only $1$, we subtract exponents instead of ever evaluating $2^{2001}$ or $3^{2003}$. Tool #3 (Eliminate Possibilities) guards the two built-in traps — subtracting the exponents the wrong way gives (D) $\tfrac23$, and dropping the leftover $3$ gives (C) $\tfrac12$.
Execute — Answer: E
8.EE.A.1 Step 1 Break the base 6 apart
- The denominator uses base $6$, which shares no base with the numerator.
- Since $6=2\cdot 3$, the power-of-a-product rule gives $6^{2002}=(2\cdot 3)^{2002}=2^{2002}\cdot 3^{2002}$.
- Now every base in the whole fraction is either $2$ or $3$.
💡 A power of a product is the product of the powers, so $6^n$ is just $2^n$ times $3^n$.
6.EE.A.3 Step 2 Group the 2s and the 3s
- Rewrite the fraction with the matching bases side by side: $\dfrac{2^{2001}\cdot 3^{2003}}{2^{2002}\cdot 3^{2002}}=\dfrac{2^{2001}}{2^{2002}}\cdot\dfrac{3^{2003}}{3^{2002}}$.
- One messy fraction has become two tidy same-base fractions that can be handled separately.
💡 Multiplication lets you shuffle the factors so like bases sit together.
8.EE.A.1 Step 3 Subtract exponents on each base
- For a quotient of equal bases you subtract exponents.
- On base $2$: $\dfrac{2^{2001}}{2^{2002}}=2^{2001-2002}=2^{-1}=\dfrac12$.
- On base $3$: $\dfrac{3^{2003}}{3^{2002}}=3^{2003-2002}=3^{1}=3$.
- The huge exponents never need to be evaluated — only their difference of $1$ matters.
💡 Dividing same-base powers just subtracts the exponents, so the gap of $1$ is all that survives.
5.NF.B.4 Step 4 Multiply the two pieces
- Combine the results: $\dfrac12\cdot 3=\dfrac{3}{2}$.
- That is choice (E).
- Check the traps: subtracting the exponents backwards would give $2^{1}\cdot 3^{-1}=\dfrac23$, which is the decoy (D); forgetting that the base-$3$ part leaves a $3$ behind gives just $\dfrac12$, the decoy (C).
- Neither survives a careful pass.
💡 Multiplying a fraction by a whole number scales its numerator.
8.EE.A.1 The denominator uses base $6$, which shares no base with the numerator. Since $6 6.EE.A.3 Rewrite the fraction with the matching bases side by side: $\dfrac{2^{2001}\cdot 8.EE.A.1 For a quotient of equal bases you subtract exponents. On base $2$: $\dfrac{2^{20 5.NF.B.4 Combine the results: $\dfrac12\cdot 3=\dfrac{3}{2}$. That is choice (E). Check t Review
Reasonableness: A quick sanity test on tiny exponents confirms the pattern: $\dfrac{2^{1}\cdot 3^{3}}{6^{2}}=\dfrac{2\cdot 27}{36}=\dfrac{54}{36}=\dfrac32$, the same value. That makes sense because the base-$2$ exponent sits one below the shared power (giving $2^{-1}$) while the base-$3$ exponent sits one above it (giving $3^{+1}$), so the answer is $\dfrac{3}{2}$ regardless of how large the shared exponent is. A value just above $1$ is reasonable, and it rules out the small fractions (A)–(D).
Alternative: Write everything over base $6$ instead of splitting: $2^{2001}\cdot 3^{2003}=2^{2001}\cdot 3^{2001}\cdot 3^{2}=6^{2001}\cdot 9$. Then $\dfrac{6^{2001}\cdot 9}{6^{2002}}=\dfrac{9}{6}=\dfrac32$, landing on (E) with a single exponent subtraction.
CCSS standards used (min grade 8)
8.EE.A.1Know and apply the properties of integer exponents (Splitting $6^{2002}=2^{2002}\cdot 3^{2002}$ with the power-of-a-product rule and subtracting exponents in each quotient, including the negative exponent $2^{-1}=\tfrac12$.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Regrouping the fraction so the like bases $2$ and $3$ sit together as two separate quotients.)5.NF.B.4Apply and extend understanding of multiplication to multiply a fraction by a fraction (Combining $\tfrac12\cdot 3=\tfrac32$ in the final step.)
⭐ Break every base down to primes so they match, then dividing same-base powers is just subtracting the exponents — the giant numbers never need to be worked out.
⭐ Break every base down to primes so they match, then dividing same-base powers is just subtracting the exponents — the giant numbers never need to be worked out.
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