AMC 10 · 2011 · #19

Grade 8 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

perfect-squaresdifference-of-squaresfactors convert-to-algebracasework ↑ Prerequisites: perfect-squares

📏 Long solution 💡 3 insights

Problem

In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town's population during this twenty-year period?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2011 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: A town's population is a perfect square in 1991. Add 150 and it becomes 9 more than a perfect square (2001). Add another 150 and it is a perfect square again (2011). Find which choice is closest to the percent growth over the whole twenty years.

Givens: In 1991 the population is a perfect square; In 2001 the population (150 more) equals 9 more than a perfect square; In 2011 the population (another 150 more) is again a perfect square; The total rise from 1991 to 2011 is $150 + 150 = 300$ people

Unknowns: The 1991 population, and from it the percent growth from 1991 to 2011

Understand

Plan

Primary tool: #4 Introduce a Variable

Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities

The clues describe three unknown populations linked by fixed jumps of 150. I name the perfect squares with letters, turn each clue into an equation, and get a difference of two squares equal to a small number. A difference of squares factors, so I can list the few factor pairs and test which one also satisfies the 2011 condition.

Execute — Answer: E

#4 Introduce a Variable 6.EE.A.2 Step 1

Name the perfect squares

Let the 1991 population be $n^2$.
In 2001 it is $n^2 + 150$, and this equals $9$ more than a perfect square, so write it as $m^2 + 9$.
In 2011 it is $n^2 + 300$, another perfect square, so call it $k^2$.

$$1991:\ n^2 \qquad 2001:\ n^2+150=m^2+9 \qquad 2011:\ n^2+300=k^2$$

💡 Giving each unknown square its own letter turns a word story into equations you can push around.

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

Turn the 2001 clue into an equation

From $n^2 + 150 = m^2 + 9$, move the numbers to one side: $m^2 - n^2 = 150 - 9 = 141$.
So the two perfect squares $m^2$ and $n^2$ differ by exactly $141$.

$$n^2 + 150 = m^2 + 9 \ \Rightarrow\ m^2 - n^2 = 141$$

💡 Collecting the plain numbers on one side leaves a clean relationship between the two squares.

#2 Make a Systematic List 4.OA.B.4 Step 3

Factor the difference of squares

A difference of two squares factors: $m^2 - n^2 = (m-n)(m+n)$.
So $(m-n)(m+n) = 141$.
Since $141 = 3 \times 47$, its factor pairs are $1 \times 141$ and $3 \times 47$.
Because $m$ and $n$ are positive, $m-n$ is the smaller factor and $m+n$ the larger.

$$(m-n)(m+n)=141=1\times141=3\times47$$

💡 Splitting a product back into its factor pairs gives only a handful of cases to test.

#4 Introduce a Variable 8.EE.C.7 Step 4

Solve for n in each case

For a pair, adding the two equations $m-n=a$ and $m+n=b$ gives $2m=a+b$, and subtracting gives $2n=b-a$.
Pair $1\times141$: $2n = 141-1 = 140$, so $n = 70$.
Pair $3\times47$: $2n = 47-3 = 44$, so $n = 22$.

$$1\times141:\ n=\tfrac{141-1}{2}=70 \qquad 3\times47:\ n=\tfrac{47-3}{2}=22$$

💡 Adding and subtracting the two simple equations peels apart $m$ and $n$ one at a time.

#3 Eliminate Possibilities 8.EE.A.2 Step 5

Keep the case that fits 2011

The 2011 population $n^2 + 300$ must also be a perfect square.
Test $n = 70$: $70^2 + 300 = 4900 + 300 = 5200$, and $72^2 = 5184$, $73^2 = 5329$, so $5200$ is not a perfect square — reject it.
Test $n = 22$: $22^2 + 300 = 484 + 300 = 784 = 28^2$, a perfect square.
So $n = 22$ is the town, with 1991 population $484$ and 2011 population $784$.

$$70^2+300=5200\ (\text{not square}) \qquad 22^2+300=784=28^2\ (\text{square})$$

💡 Only the case that satisfies both perfect-square clues at once can be the real town.

#4 Introduce a Variable 7.RP.A.3 Step 6

Compute the percent growth

The population rose from $484$ in 1991 to $784$ in 2011, a rise of $300$.
Percent growth is the rise divided by the start: $\dfrac{300}{484} \approx 0.620 = 62\%$.
The closest choice is $62$, which is (E).

$$\frac{784-484}{484}=\frac{300}{484}\approx0.620=62\%$$

💡 Percent growth compares how much you gained to what you started with, not to where you ended.

[1] #4 6.EE.A.2 Let the 1991 population be $n^2$. In 2001 it is $n^2 + 150$, and this equals $9$

[2] #4 7.EE.B.4 From $n^2 + 150 = m^2 + 9$, move the numbers to one side: $m^2 - n^2 = 150 - 9 =

[3] #2 4.OA.B.4 A difference of two squares factors: $m^2 - n^2 = (m-n)(m+n)$. So $(m-n)(m+n) =

[4] #4 8.EE.C.7 For a pair, adding the two equations $m-n=a$ and $m+n=b$ gives $2m=a+b$, and sub

[5] #3 8.EE.A.2 The 2011 population $n^2 + 300$ must also be a perfect square. Test $n = 70$: $7

[6] #4 7.RP.A.3 The population rose from $484$ in 1991 to $784$ in 2011, a rise of $300$. Percen

Review

Reasonableness: Check all three clues with $n=22$: 1991 is $484=22^2$ (square), 2001 is $634 = 625 + 9 = 25^2 + 9$ (9 more than a square), 2011 is $784=28^2$ (square). All fit. The growth $300/484$ is a bit under two-thirds, and $62\%$ is the only choice near that, so (E) is sound.

Alternative: Since the choices sit near $60\%$, the 1991 population must be near $300 / 0.6 \approx 500$. Scan perfect squares close to $500$: $22^2 = 484$ and $28^2 = 784$ differ by exactly $300$, and $484 + 150 = 634 = 25^2 + 9$ checks the middle clue. This lands on the same town without factoring.

CCSS standards used (min grade 8)

6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Naming the three unknown populations as $n^2$, $m^2+9$, and $k^2$.)
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities (Turning the 2001 clue into the equation $m^2 - n^2 = 141$.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Listing the factor pairs of $141$ after writing it as $(m-n)(m+n)$.)
8.EE.C.7 Solve linear equations in one variable (Adding and subtracting $m-n=a$ and $m+n=b$ to solve for $n$ in each case.)
8.EE.A.2 Use square root and cube root symbols to represent solutions (Testing whether $n^2+300$ is a perfect square to pick the right case.)
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems (Dividing the rise of 300 by the starting population 484 to get the percent growth.)

⭐ When two perfect squares differ by a small number, factor that number as $(m-n)(m+n)$ to find the squares, then measure growth against where you started.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Name the perfect squares

Turn the 2001 clue into an equation

Factor the difference of squares

Solve for n in each case

Keep the case that fits 2011

Compute the percent growth

Review

CCSS standards used (min grade 8)

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