AMC 10 · 2002 · #20

Grade 8 number-theory

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

fraction-decimal-conversiongcdprime-factorization systematic-enumeration ↑ Prerequisites: fraction-decimal-conversiongcd

📏 Long solution 💡 3 insights

Problem

A repeating decimal 0.ab is built from two digits a and b that are not both 9 and not both 0. Write it as a fraction in lowest terms and count how many different denominators can appear.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Introduce a Variable

Testing all 98 allowed digit pairs one at a time would work but wastes the structure. Tool #4 (Introduce a Variable) names the repeating block N=10a+b, and tool #13 (Convert to Algebra) collapses the endless decimal into the single fraction N/99 by shifting and subtracting. Tool #16 (Change Focus) then moves the question off the fraction and onto gcd(N,99): reducing divides the denominator by that gcd, so the denominator is always 99/gcd(N,99) and can only ever be a divisor of 99. Tool #2 (Make a Systematic List) lists those divisors, which gives an upper bound and not yet an answer. Tool #3 (Eliminate Possibilities) knocks out the one divisor the bans forbid, and the final, most-skipped move is tool #6 (Guess and Check): exhibit a concrete block for every divisor that survives, so the count is proved rather than assumed.

1STEP 1

Name the repeating block

Name the block N; the two bans remove N=99 and N=0, leaving 1 ≤ N ≤ 98.

N=10a+b, 1 ≤ N ≤ 98

2STEP 2

Turn the decimal into a fraction

Multiplying by 100 and subtracting cancels the tail: 99x = N, so x = N/99.

100x=N+x → 99x=N → x=N/99

3STEP 3

Reduce, and watch the denominator

Reducing divides both by the gcd, so the denominator is always 99/g, a divisor of 99.

N/99=(N/g)/(99/g), g=gcd(N,99), denominator=99/g

4STEP 4

List the divisors of 99

Since 99 = 3² · 11, the divisors are 1, 3, 9, 11, 33, 99 — an upper bound of six.

99=3² · 11, divisors = {1,3,9,11,33,99}

5STEP 5

Rule out the denominator 1

Denominator 1 needs 99 to divide N, impossible in range — that is exactly what the two bans remove.

99/g=1 ⇔ 99 ∣ N ⇔ N∈{0,99} (both banned)

6STEP 6

Show the other five all occur

Blocks 01, 03, 09, 11, 33 realise the other five, so the count is 5, choice (C).

1/99, 1/33, 1/11, 1/9, 1/3 → 5 denominators → (C)

Answer

The count comes out one below the number of divisors of 99, which is the shape to expect: reduction can reach any divisor, and only the extreme case — the fraction collapsing to a whole number — is blocked, which is precisely what the problem's two bans block. A random spot-check agrees with the list: the block 45 gives 45/99=5/11, denominator 11, already accounted for. The distractors are all counting the wrong objects: (E) 9 counts digits rather than denominators, (D) 8 counts something larger than the six divisors that exist at all, and (B) 4 is what you get by dropping one of the five — most easily 99 itself, which is achieved only by blocks like 01 that look degenerate but are perfectly legal.

💡Key takeaway

Every two-digit repeating decimal is something over 99, so its reduced denominator has to be a divisor of 99 — then check one by one which of those divisors really show up.

Name the repeating block
Turn the decimal into a fraction
Reduce, and watch the denominator
List the divisors of 99
Rule out the denominator 1
Show the other five all occur