AMC 10 · 2007 · #17
Grade 8 arithmeticSuppose that m and n are positive integers such that 75m=n3. What is the minimum possible value of m+n?
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: You want two positive whole numbers $m$ and $n$ with $75m = n^3$, meaning $75$ times $m$ lands exactly on a perfect cube. Among all such pairs, find the smallest possible value of $m + n$.
Givens: $m$ and $n$ are positive integers; $75m = n^3$, so $75m$ must be a perfect cube; Answer choices: (A) $15$, (B) $30$, (C) $50$, (D) $60$, (E) $5700$
Unknowns: The smallest possible value of $m + n$
Understand
Restated: You want two positive whole numbers $m$ and $n$ with $75m = n^3$, meaning $75$ times $m$ lands exactly on a perfect cube. Among all such pairs, find the smallest possible value of $m + n$.
Givens: $m$ and $n$ are positive integers; $75m = n^3$, so $75m$ must be a perfect cube; Answer choices: (A) $15$, (B) $30$, (C) $50$, (D) $60$, (E) $5700$
Plan
Primary tool: #14 Extreme Principle
Secondary: #7 Identify Subproblems, #4 Introduce a Variable
The words "minimum possible value" point straight at Tool #14 (Extreme Principle): we push $m$ and $n$ down to the smallest sizes the cube condition allows. To see what that condition really is, Tool #7 (Identify Subproblems) says: first break $75$ into primes, then handle each prime's exponent on its own. Tool #4 (Introduce a Variable) writes $n = 3^x \cdot 5^y$ so the cube $n^3$ has exponents $3x$ and $3y$ we can control. Matching those tripled exponents against what $75$ already supplies tells us the smallest legal $x$ and $y$, which pins down the smallest $n$, then the smallest $m$, then their sum.
Execute — Answer: D
4.OA.B.4 Step 1 Break 75 into primes
- Factor $75$ into prime building blocks so we can see exactly what $75m$ starts with.
- $75 = 3 \cdot 25 = 3 \cdot 5^2$.
- So $75$ carries one factor of $3$ and two factors of $5$.
- Everything about being a cube will come down to those two exponents, $1$ and $2$.
💡 Primes are the atoms of a number, so factoring first shows exactly which pieces $m$ still has to supply.
8.EE.A.1 Step 2 Write the cube in prime form
- Since $75m = n^3$ and $75$ is made only of the primes $3$ and $5$, the cube $n^3$ can contain no other primes, which means $n$ itself is built only from $3$ and $5$.
- Write $n = 3^{x} \cdot 5^{y}$ for whole numbers $x, y \ge 0$.
- Cubing multiplies each exponent by $3$: $n^3 = 3^{3x} \cdot 5^{3y}$.
- A perfect cube is exactly a number whose every prime exponent is a multiple of $3$, so this is the shape $75m$ must match.
💡 Cubing triples every exponent, so a cube is just a number where all the prime powers are multiples of three.
8.EE.A.2 Step 3 Push the exponents to their minimum
- The cube $n^3 = 3^{3x} \cdot 5^{3y}$ must be a multiple of $75 = 3^{1} \cdot 5^{2}$, so its exponents can't fall short: we need $3x \ge 1$ and $3y \ge 2$.
- Because $3x$ and $3y$ are multiples of $3$, the smallest allowed values are $3x = 3$ (so $x = 1$) and $3y = 3$ (so $y = 1$).
- That gives the smallest $n = 3^{1} \cdot 5^{1} = 15$, and its cube is $n^3 = 3^{3} \cdot 5^{3} = 27 \cdot 125 = 3375$, which is indeed the perfect cube $15^3$.
💡 The smallest multiple of three that still clears each hurdle is three itself, so both exponents bottom out at one.
4.NBT.B.6 Step 4 Recover m and add
- Now find $m$ from $75m = n^3 = 3375$: divide, $m = 3375 \div 75 = 45$.
- Since we already made $n$ as small as possible and $m$ is forced by that choice, this is the smallest legal pair.
- Add them: $m + n = 45 + 15 = 60$.
- That is choice (D).
💡 Once the cube is fixed at its smallest, $m$ is whatever is left after dividing out the $75$, and the sum follows.
4.OA.B.4 Factor $75$ into prime building blocks so we can see exactly what $75m$ starts w 8.EE.A.1 Since $75m = n^3$ and $75$ is made only of the primes $3$ and $5$, the cube $n^3 8.EE.A.2 The cube $n^3 = 3^{3x} \cdot 5^{3y}$ must be a multiple of $75 = 3^{1} \cdot 5^{ 4.NBT.B.6 Now find $m$ from $75m = n^3 = 3375$: divide, $m = 3375 \div 75 = 45$. Since we Review
Reasonableness: Check the pair really works: $75 \cdot 45 = 3375$ and $15^3 = 15 \cdot 15 \cdot 15 = 225 \cdot 15 = 3375$, so $75m = n^3$ holds exactly. It is also genuinely the minimum: any smaller cube multiple of $75$ would need exponents below $3$ on the $3$ or the $5$, but a cube can't have an exponent of $1$ or $2$, so $15^3$ is the first cube that $75$ divides. The answer $60$ sits sensibly among the choices: (A) $15$ and (B) $30$ are too small (even $n = 15$ alone already uses up $15$, and $m$ can't be $0$), while (E) $5700$ comes from wastefully overshooting the exponents.
Alternative: Skip the algebra and list perfect cubes until one is a multiple of $75$: $1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375$. Test divisibility by $75$; the first that works is $3375 = 15^3$, giving $n = 15$ and $m = 3375 / 75 = 45$, so $m + n = 60$. The prime-factor argument just tells you in advance that $15^3$ is where the search must land.
CCSS standards used (min grade 8)
4.OA.B.4Find all factor pairs for a whole number and recognize prime factorization (Breaking $75$ into its prime factors $3^{1} \cdot 5^{2}$ so the cube condition can be read off the exponents.)8.EE.A.1Know and apply the properties of integer exponents (Cubing $n = 3^{x} \cdot 5^{y}$ into $n^{3} = 3^{3x} \cdot 5^{3y}$ to see that a cube's exponents are all multiples of $3$.)8.EE.A.2Use cube root symbols and evaluate cube roots of perfect cubes (Choosing the smallest exponents so $n = 15$ and confirming $n^3 = 15^3 = 3375$ is a perfect cube.)4.NBT.B.6Find whole-number quotients using place value and the properties of operations (Dividing $3375 \div 75 = 45$ to recover $m$, then adding to get $m + n = 60$.)
⭐ To make a product a perfect cube, factor into primes and bump each exponent up to the nearest multiple of three, no higher.
⭐ To make a product a perfect cube, factor into primes and bump each exponent up to the nearest multiple of three, no higher.
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