AMC 10 · 2012 · #17

Grade 7 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

difference-of-cubesgcddivisibility-rules convert-to-algebracasework ↑ Prerequisites: prime-factorization

📏 Medium solution 💡 3 insights

Problem

Let $a$ and $b$ be relatively prime positive integers with $a>b>0$ and $\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}$ . What is $a-b$ ?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2012 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.

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Toolkit + CCSS Solution

Understand

Restated: Two positive whole numbers a and b share no common factor bigger than 1, with a larger than b. The value of (a cubed minus b cubed) divided by (a minus b) cubed equals 73/3. Find the value of a minus b.

Givens: a and b are positive integers with a > b > 0.; a and b are relatively prime (their greatest common factor is 1).; (a^3 - b^3) / (a - b)^3 = 73/3.

Unknowns: The value of the difference a - b, which is one of 1, 2, 3, 4, 5.

Understand

Givens: a and b are positive integers with a > b > 0.; a and b are relatively prime (their greatest common factor is 1).; (a^3 - b^3) / (a - b)^3 = 73/3.

Plan

Primary tool: #3 Eliminate Possibilities

Secondary: #7 Identify Subproblems, #4 Introduce a Variable, #13 Convert to Algebra

The answer is one of five whole numbers, so the fastest route is to turn the messy fraction into a clean divisibility fact about a - b, then throw out every choice that fails it. Factoring the difference of cubes collapses the ratio, and a short algebra step turns it into 9ab = 70(a - b)^2. That single equation tells us a - b must be a multiple of 3, and only one choice qualifies.

Execute — Answer: C

#7 Identify Subproblems 6.EE.A.3 Step 1

Factor the difference of cubes

The top is a difference of cubes: a^3 - b^3 = (a - b)(a^2 + ab + b^2).
The bottom is (a - b)^3.
Cancel one factor of (a - b) from top and bottom.
The whole fraction becomes (a^2 + ab + b^2) over (a - b)^2.

$$\dfrac{a^3-b^3}{(a-b)^3}=\dfrac{(a-b)(a^2+ab+b^2)}{(a-b)^3}=\dfrac{a^2+ab+b^2}{(a-b)^2}$$

💡 Factoring the top exposes a shared (a - b) that cancels, shrinking a cubic ratio down to a squared one.

#4 Introduce a Variable 7.EE.B.4 Step 2

Split off the whole-number part

Rewrite the top using a^2 + ab + b^2 = (a - b)^2 + 3ab.
Dividing by (a - b)^2 splits the fraction into 1 plus 3ab over (a - b)^2.
Set that equal to 73/3.

$$a^2+ab+b^2=(a-b)^2+3ab\ \Rightarrow\ 1+\dfrac{3ab}{(a-b)^2}=\dfrac{73}{3}$$

💡 Pulling out the built-in 1 leaves a single clean fraction to pin down.

#13 Convert to Algebra 7.EE.B.4 Step 3

Clear the fractions

Subtract 1 from both sides: 73/3 minus 1 is 70/3, so 3ab over (a - b)^2 equals 70/3.
Cross-multiply to remove all fractions.
The 3 on the left cancels into the 3 on the right, leaving 9ab = 70(a - b)^2.

$$\dfrac{3ab}{(a-b)^2}=\dfrac{70}{3}\ \Rightarrow\ 9ab=70(a-b)^2$$

💡 Clearing fractions turns a ratio condition into one tidy whole-number equation.

#3 Eliminate Possibilities 6.NS.B.4 Step 4

Force a - b to be a multiple of 3

The left side, 9ab, is divisible by 9.
So the right side, 70(a - b)^2, must also be divisible by 9.
But 70 shares no factor with 9, so all of that divisibility has to come from (a - b)^2.
If 9 divides (a - b)^2, then 3 must divide (a - b).
So a - b is a multiple of 3.

$$9\mid 70(a-b)^2,\ \gcd(70,9)=1\ \Rightarrow\ 9\mid (a-b)^2\ \Rightarrow\ 3\mid (a-b)$$

💡 A factor of 9 on one side has to be matched on the other, and 70 cannot supply it, so a - b must carry the 3.

#3 Eliminate Possibilities 6.NS.B.4 Step 5

Eliminate and check the survivor

Among the choices 1, 2, 3, 4, 5, the only multiple of 3 is 3, so every other choice is out.
Confirm it is real: with a - b = 3, the equation 9ab = 70 times 9 gives ab = 70, and a - b = 3 with ab = 70 yields a = 10, b = 7.
They are relatively prime and give 219/9 = 73/3.
So a - b = 3, answer (C).

$$a-b=3:\ 9ab=70\cdot 9\Rightarrow ab=70,\ a=10,\ b=7,\ \gcd(10,7)=1$$

💡 Once only one choice can possibly work, a quick construction confirms it really does.

[1] #7 6.EE.A.3 The top is a difference of cubes: a^3 - b^3 = (a - b)(a^2 + ab + b^2). The botto

[2] #4 7.EE.B.4 Rewrite the top using a^2 + ab + b^2 = (a - b)^2 + 3ab. Dividing by (a - b)^2 sp

[3] #13 7.EE.B.4 Subtract 1 from both sides: 73/3 minus 1 is 70/3, so 3ab over (a - b)^2 equals 7

[4] #3 6.NS.B.4 The left side, 9ab, is divisible by 9. So the right side, 70(a - b)^2, must also

[5] #3 6.NS.B.4 Among the choices 1, 2, 3, 4, 5, the only multiple of 3 is 3, so every other cho

Review

Reasonableness: Plug a = 10, b = 7 back in: a^3 - b^3 = 1000 - 343 = 657, and (a - b)^3 = 27, so the ratio is 657/27 = 73/3, exactly as required. Also a = 10 and b = 7 are relatively prime and a > b > 0, so every condition holds and a - b = 3 is confirmed.

Alternative: Skip the elimination and solve directly: from 9ab = 70(a - b)^2, write s = a + b and d = a - b so that ab = (s^2 - d^2)/4. Substituting gives 9(s^2 - d^2) = 280 d^2, so 9 s^2 = 289 d^2, meaning 3s = 17d. With gcd(a, b) = 1 the smallest solution is d = 3, s = 17, giving a = 10, b = 7 again.

CCSS standards used (min grade 7)

6.EE.A.3 Apply the properties of operations to generate equivalent expressions (Factoring a^3 - b^3 as (a - b)(a^2 + ab + b^2) and cancelling to reduce the ratio to (a^2 + ab + b^2)/(a - b)^2.)
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities (Rewriting the ratio as 1 + 3ab/(a - b)^2 = 73/3 and clearing fractions to reach 9ab = 70(a - b)^2.)
6.NS.B.4 Find greatest common factor and least common multiple of two numbers (Using that 70 and 9 share no common factor to conclude 9 divides (a - b)^2, hence 3 divides a - b, and checking that a = 10, b = 7 are relatively prime.)

⭐ Factor the difference of cubes, boil the fraction down to one equation, and let a divisibility rule knock out every answer but the one that fits.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Factor the difference of cubes

Split off the whole-number part

Clear the fractions

Force a - b to be a multiple of 3

Eliminate and check the survivor

Review

CCSS standards used (min grade 7)

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