AMC 10 · 2002 · #16

Grade 8 number-theory

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

perfect-squaresdivisibility-rulesfactors bound-inequality-then-enumerate ↑ Prerequisites: perfect-squaresdivisibility-rules

📏 Long solution 💡 3 insights

Problem

For how many integers $n$ is $\dfrac n{20-n}$ the square of an integer?

Pick an answer.

(A)

$\ 1$

(B)

$\ 2$

(C)

$\ 3$

(D)

$\ 4$

(E)

$\ 10$

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.

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Toolkit + CCSS Solution

Understand

Restated: Count how many integers n make the value of n divided by (20 minus n) equal to the square of an integer, that is, a perfect square like 0, 1, 4, or 9.

Givens: n is an integer; The expression is n / (20 - n); Its value has to be a perfect square (the square of some integer)

Unknowns: How many integers n make n / (20 - n) a perfect square

Understand

Restated: Count how many integers n make the value of n divided by (20 minus n) equal to the square of an integer, that is, a perfect square like 0, 1, 4, or 9.

Givens: n is an integer; The expression is n / (20 - n); Its value has to be a perfect square (the square of some integer)

Plan

Primary tool: #4 Introduce a Variable

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

Instead of testing many values of n, name the square: let k be the integer whose square equals the fraction. Solving for n turns the problem into a divisibility question, and then a short list of the divisors of 20 gives every answer at once.

Execute — Answer: D

#3 Eliminate Possibilities 7.NS.A.2 Step 1

Where the fraction can be a square

A perfect square is never negative, so the fraction must be 0 or positive.
The numerator n and the denominator 20 - n must therefore share the same sign (or n = 0).
If n > 20 the numerator is positive but 20 - n is negative, giving a negative value.
If n < 0 the numerator is negative while 20 - n is positive, again negative.
And n = 20 is not allowed.
So the only integers worth checking are 0 through 19.

$$\dfrac{n}{20-n}\ge 0 \;\Rightarrow\; 0 \le n \le 19$$

💡 A square can't be negative, so the top and bottom of the fraction must pull in the same direction.

#4 Introduce a Variable 8.EE.C.7 Step 2

Name the square and solve for n

Let k be the integer whose square is the fraction, so k squared equals n over (20 - n).
Multiply both sides by 20 - n to clear the fraction, then gather every n on one side.
This gives n times (1 + k squared) equal to 20 times k squared.

$$k^2=\dfrac{n}{20-n} \;\Rightarrow\; n = k^2(20-n) \;\Rightarrow\; n(1+k^2)=20k^2$$

💡 Giving the square a name, k, lets you rearrange the fraction into a clean equation for n.

#4 Introduce a Variable 4.OA.B.4 Step 3

Turn it into a divisibility test

Solve for n: n equals 20 times k squared, divided by (1 + k squared).
Rewrite the top as 20 times (1 + k squared) minus 20, so n = 20 minus 20 over (1 + k squared).
For n to be an integer, 1 + k squared must divide 20 exactly.

$$n=\dfrac{20k^2}{1+k^2}=20-\dfrac{20}{1+k^2} \;\Rightarrow\; (1+k^2)\mid 20$$

💡 Splitting off the remainder shows n is whole exactly when 1 + k squared is a divisor of 20.

#2 Make a Systematic List 6.EE.A.1 Step 4

List the divisors and keep the squares

The positive divisors of 20 are 1, 2, 4, 5, 10, 20.
Setting 1 + k squared equal to each one means k squared is 0, 1, 3, 4, 9, or 19.
Of these, only 0, 1, 4, and 9 are actually squares of an integer, giving k = 0, 1, 2, 3.
The values 3 and 19 are not perfect squares, so they are thrown out.

$$1+k^2\in\{1,2,4,5,10,20\} \Rightarrow k^2\in\{0,1,4,9\} \Rightarrow k=0,1,2,3$$

💡 Only divisors of 20 that are one more than a perfect square can come from a real integer k.

#3 Eliminate Possibilities 4.OA.B.4 Step 5

Read off n and count

Plug each k back into n = 20 minus 20 over (1 + k squared): k = 0 gives n = 0, k = 1 gives n = 10, k = 2 gives n = 16, k = 3 gives n = 18.
Each result sits inside 0 to 19, and each fraction is indeed a square: 0, 1, 4, 9.
That is four integers, so the answer is (D).

$$k=0{:}\;n=0,\; k=1{:}\;n=10,\; k=2{:}\;n=16,\; k=3{:}\;n=18$$

💡 Each surviving k feeds back one valid n, and there are exactly four of them.

[1] #3 7.NS.A.2 A perfect square is never negative, so the fraction must be 0 or positive. The n

[2] #4 8.EE.C.7 Let k be the integer whose square is the fraction, so k squared equals n over (2

[3] #4 4.OA.B.4 Solve for n: n equals 20 times k squared, divided by (1 + k squared). Rewrite th

[4] #2 6.EE.A.1 The positive divisors of 20 are 1, 2, 4, 5, 10, 20. Setting 1 + k squared equal

[5] #3 4.OA.B.4 Plug each k back into n = 20 minus 20 over (1 + k squared): k = 0 gives n = 0, k

Review

Reasonableness: Check each value directly: 0/(20-0)=0=0^2, 10/(20-10)=1=1^2, 16/(20-16)=4=2^2, 18/(20-18)=9=3^2. All four are perfect squares, and no other n between 0 and 19 works because every valid n must make 1 + k squared a divisor of 20. So there are 4 integers, matching choice (D).

Alternative: Skip the algebra and reason about size. For n from 1 to 9 the fraction is between 0 and 1, so it can never be a whole square except by being 0 (only n = 0 does that). For n from 10 to 19, test directly: n = 10, 16, 18 give 1, 4, 9, while the rest give non-squares. Adding n = 0, that is again 4 integers.

CCSS standards used (min grade 8)

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers (Using signs of the numerator and denominator to see the fraction is non-negative only for 0 <= n <= 19)
8.EE.C.7 Solve linear equations in one variable (Clearing the fraction and collecting n to get n(1 + k^2) = 20k^2)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Requiring 1 + k^2 to divide 20 and listing the divisors of 20)
6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents (Recognizing which values of k^2 are genuine perfect squares (0, 1, 4, 9))

⭐ Rewrite the fraction as n = 20 minus 20/(k squared + 1); then k squared + 1 only has to be a divisor of 20, which happens just for k = 0, 1, 2, 3, giving four values of n.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Where the fraction can be a square

Name the square and solve for n

Turn it into a divisibility test

List the divisors and keep the squares

Read off n and count

Review

CCSS standards used (min grade 8)

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