AMC 10 · 2007 · #9
Grade 8 arithmeticReal numbers a and b satisfy the equations 3a=81b+2 and 125b=5a−3. What is ab?
Pick an answer.
AMC 10 2007 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: Two equations tie together the real numbers $a$ and $b$: $3^{a} = 81^{b+2}$ and $125^{b} = 5^{a-3}$. Find the value of the product $ab$.
Givens: The equation $3^{a} = 81^{b+2}$; The equation $125^{b} = 5^{a-3}$; $a$ and $b$ are real numbers; Answer choices: (A) $-60$, (B) $-17$, (C) $9$, (D) $12$, (E) $60$
Unknowns: The value of the product $ab$
Understand
Restated: Two equations tie together the real numbers $a$ and $b$: $3^{a} = 81^{b+2}$ and $125^{b} = 5^{a-3}$. Find the value of the product $ab$.
Givens: The equation $3^{a} = 81^{b+2}$; The equation $125^{b} = 5^{a-3}$; $a$ and $b$ are real numbers; Answer choices: (A) $-60$, (B) $-17$, (C) $9$, (D) $12$, (E) $60$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #5 Look for a Pattern, #3 Eliminate Possibilities
Each equation mixes two different-looking bases, so Tool #7 (Identify Subproblems) treats them one at a time. The key that unlocks both is Tool #5 (Look for a Pattern): $81$ and $125$ are secretly powers of $3$ and $5$, so every term can be written on a single base. Once the bases match, equal powers force equal exponents, and each equation collapses into a plain linear relation between $a$ and $b$. Two linear relations make a system that pins down $a$ and $b$, and then $ab$ is one multiplication. Tool #3 (Eliminate Possibilities) offers a fast sign check at the end.
Execute — Answer: E
8.EE.A.1 Step 1 Write every base as a small prime power
- The two sides of each equation use different bases, which blocks any direct comparison.
- But $81 = 3 \times 3 \times 3 \times 3 = 3^{4}$ and $125 = 5 \times 5 \times 5 = 5^{3}$.
- Rewriting the big bases this way puts each equation on one common base — base $3$ for the first, base $5$ for the second.
💡 A messy base is often a familiar small base in disguise, so rewrite it and the two sides finally speak the same language.
8.EE.A.1 Step 2 Match exponents in the first equation
- Replace $81$ with $3^{4}$: the first equation becomes $3^{a} = (3^{4})^{b+2} = 3^{4(b+2)}$.
- A power of a power multiplies the exponents.
- Now both sides are powers of $3$, so the exponents must be equal: $a = 4(b+2) = 4b + 8$.
💡 Same base on both sides means the exponents alone carry the equation, so just set them equal.
8.EE.A.1 Step 3 Match exponents in the second equation
- Replace $125$ with $5^{3}$: the second equation becomes $(5^{3})^{b} = 5^{3b}$ on the left and $5^{a-3}$ on the right.
- Both sides are powers of $5$, so their exponents are equal: $3b = a - 3$, which rearranges to $a = 3b + 3$.
💡 Once the base is shared, an exponential statement turns into an ordinary linear one.
8.EE.C.8 Step 4 Combine the two relations
- Now $a$ equals two things at once: $a = 4b + 8$ and $a = 3b + 3$.
- Since both describe the same $a$, set the right-hand sides equal to each other.
- This eliminates $a$ and leaves a single equation in $b$: $4b + 8 = 3b + 3$.
💡 Two formulas for the same quantity can be set equal, which cancels that quantity and isolates the other unknown.
8.EE.C.7 Step 5 Solve for b, then a
- Solve $4b + 8 = 3b + 3$.
- Subtract $3b$ from both sides: $b + 8 = 3$.
- Subtract $8$: $b = -5$.
- Put $b = -5$ back into $a = 3b + 3$: $a = 3(-5) + 3 = -15 + 3 = -12$.
💡 Peel one variable off a step at a time, then feed its value back to unlock the other.
7.NS.A.2 Step 6 Multiply to get ab
- The question asks for $ab$, so multiply the two values: $ab = (-12)(-5)$.
- A negative times a negative is positive, and $12 \times 5 = 60$, so $ab = 60$.
- That matches choice (E).
💡 Two negatives multiplied cancel their signs, leaving a positive product.
8.EE.A.1 The two sides of each equation use different bases, which blocks any direct comp 8.EE.A.1 Replace $81$ with $3^{4}$: the first equation becomes $3^{a} = (3^{4})^{b+2} = 3 8.EE.A.1 Replace $125$ with $5^{3}$: the second equation becomes $(5^{3})^{b} = 5^{3b}$ o 8.EE.C.8 Now $a$ equals two things at once: $a = 4b + 8$ and $a = 3b + 3$. Since both des 8.EE.C.7 Solve $4b + 8 = 3b + 3$. Subtract $3b$ from both sides: $b + 8 = 3$. Subtract $8 7.NS.A.2 The question asks for $ab$, so multiply the two values: $ab = (-12)(-5)$. A nega Review
Reasonableness: Check the pair $a = -12$, $b = -5$ against both original equations. First: $4b + 8 = 4(-5) + 8 = -12 = a$, so $3^{a} = 3^{-12}$ and $81^{b+2} = 81^{-3} = 3^{-12}$ — they agree. Second: $3b = -15$ and $a - 3 = -15$, so $125^{b} = 5^{-15} = 5^{a-3}$ — they agree too. The product $(-12)(-5) = 60$ is positive, and a quick sign check rules out the negative choices (A) and (B) as well as the small positives (C) and (D), leaving (E).
Alternative: Use Eliminate Possibilities on the sign first. Both $a$ and $b$ come out negative from the linear relations, so their product must be positive — that erases (A) $-60$ and (B) $-17$ immediately. Among the positive choices, only $60$ is large enough to come from multiplying numbers near $-12$ and $-5$, which points straight to (E) even before the exact multiplication.
CCSS standards used (min grade 8)
8.EE.A.1Know and apply the properties of integer exponents (Rewriting $81 = 3^{4}$ and $125 = 5^{3}$, using the power-of-a-power rule, and matching exponents once the bases are equal.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Combining $a = 4b + 8$ and $a = 3b + 3$ into one equation by setting the two expressions for $a$ equal.)8.EE.C.7Solve linear equations in one variable (Solving $4b + 8 = 3b + 3$ for $b = -5$ and back-substituting to find $a = -12$.)7.NS.A.2Apply and extend understanding of multiplication and division of rational numbers (Multiplying the two negative values, $(-12)(-5) = 60$, to get the product $ab$.)
⭐ When an equation hides different bases, rewrite them as powers of the same small number — then equal powers mean equal exponents, and the hard exponent problem becomes easy algebra.
⭐ When an equation hides different bases, rewrite them as powers of the same small number — then equal powers mean equal exponents, and the hard exponent problem becomes easy algebra.
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