AMC 10 · 2013 · #10

Grade 8 number-theory
fraction-decimal-conversionfactorsdigit-decomposition convert-to-algebrasystematic-enumeration ↑ Prerequisites: fraction-decimal-conversion
📏 Medium solution 💡 2 insights
Problem
A unit fraction must equal a two-digit repeating decimal whose digits differ. Add every denominator that works.

Pick an answer.

(A)
11
(B)
44
(C)
110
(D)
143
(E)
155
How to solve
Strategy Introduce a Variable

An endless decimal is hard to reason about directly, so name it (Tool #4): let x = 0.ababab… and let N stand for the two-digit block value 10a+b. Multiplying x by 100 slides the decimal exactly one block over and leaves the same infinite tail, so subtracting kills the tail and turns the decimal into the fraction N/99. Setting that equal to 1/n gives nN = 99, which turns an infinite-decimal question into a finite factoring question. From there, list all factor pairs of 99 (Tool #2) and cross off the ones whose block has a = b (Tool #3). The listing must run in both directions: finding n ∣ 99 only says which n could work, so each survivor still has to be checked to make sure its digits really are two different digits.

1STEP 1

Name the decimal and its block

The repeating block is that block over 99.

100x = N + x → 99x = N → x = N/99, N = 10a+b
2STEP 2

Turn the equation into a factor pair

The equation becomes a plain factor pair.

1/n = N/99 ⟺ nN = 99, 1 ≤ N ≤ 99
3STEP 3

List every factor pair of 99

There are only six such pairs.

n & 1 & 3 & 9 & 11 & 33 & 99 ; N=99/n & 99 & 33 & 11 & 9 & 3 & 1 ; ab & 99 & 33 & 11 & 09 & 03 & 01
4STEP 4

Cross off the blocks with equal digits

The digit condition rejects half of them.

99, 33, 11 → a=b (rejected); 09, 03, 01 → a=0 ≠ b (kept)
5STEP 5

Confirm the survivors and add

The survivors add to 143, choice (D).

S = {11, 33, 99}, 11 + 33 + 99 = 143 → (D)
Answer
143
Independent count check: the sum of all divisors of 99 is (1+3+9)(1+11) = 13 · 12 = 156, and the three rejected values 1, 3, 9 add to 13, so the kept sum is 156 - 13 = 143 — matching. Structural check: nN = 99 with N ≤ 9 forces n ≥ 11, so every member of S should be a multiple of 11, and 11, 33, 99 all are. The answer choices also line up with the divisor picture, each distractor being a different slice of the same six-item list: 11 keeps only one survivor, 44 = 11+33 drops 99, 110 = 11+99 drops 33, and 155 = 156-1 is what you get by throwing out only n=1 while wrongly keeping n=3 and n=9. Only 143 corresponds to the correct rule. One ambiguity is worth settling: if the problem demanded the block be the shortest repeating block, nothing changes, since 1/11, 1/33, 1/99 each have minimal period exactly 2, while 1/3 and 1/9 have period 1 and would be excluded anyway. Both readings give the same S.
💡Key takeaway

Any decimal that repeats a two-digit block is that block over 99, so 1/n = 0.ab just means n times the block equals 99 — and asking for two different digits only throws out the blocks 11, 33, 99.

  • Name the decimal and its block
  • Turn the equation into a factor pair
  • List every factor pair of 99
  • Cross off the blocks with equal digits
  • Confirm the survivors and add