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

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.

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.

How to solve

Strategy Introduce a Variable

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.

1STEP 1

Where the fraction can be a square

A square is never negative, so n and 20 - n must share a sign — only n from 0 to 19 is worth checking.

n/(20-n) ≥ 0 → 0 ≤ n ≤ 19

2STEP 2

Name the square and solve for n

Name the square: k² = n/(20-n). Clear the denominator and gather every n to get n(1+k²) = 20k².

k²=n/(20-n) → n = k²(20-n) → n(1+k²)=20k²

3STEP 3

Turn it into a divisibility test

Solving gives n = 20k²/(1+k²) = 20 - 20/(1+k²), so n is a whole number exactly when 1+k² divides 20.

n=20k²/(1+k²)=20-20/(1+k²) → (1+k²) ∣ 20

4STEP 4

List the divisors and keep the squares

Divisors 1, 2, 4, 5, 10, 20 make k² equal 0, 1, 3, 4, 9 or 19; keep only real squares and k = 0, 1, 2, 3 survive.

1+k²∈{1,2,4,5,10,20} → k²∈{0,1,4,9} → k=0,1,2,3

5STEP 5

Read off n and count

Each k feeds back n = 0, 10, 16, 18, all in range with fractions 0, 1, 4, 9 — four integers, choice (D).

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

Answer

Check each value directly: 0/(20-0)=0=0², 10/(20-10)=1=1², 16/(20-16)=4=2², 18/(20-18)=9=3². 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).

💡Key takeaway

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.

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

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

More like this