AMC 10 · 2014 · #23

Grade 8 number-theorypattern
decimal-arithmeticplace-valuedigit-sumgeometric-series-infinite easier-related-problempattern-recognition ↑ Prerequisites: decimal-arithmeticplace-value
📏 Long solution 💡 4 insights
Problem
A fraction repeats forever and both its digits and its true period must be settled. Add the digits of one period.

Pick an answer.

(A)
874
(B)
883
(C)
887
(D)
891
(E)
892
How to solve
Strategy Solve an Easier Related Problem

Dividing 1 by 9801 by hand for two hundred digits is not a plan. The way in is 1/9²=1/81, the same problem with one-digit chunks instead of two: it expands as 0.012345679, counting 0,1,2,… and skipping one value near the end. That suggests the two-digit answer, 00,01,02,…,97,99 with 98 skipped. A guess is not a solution, though, and this is exactly where the naive counting argument leaks: writing the numbers 0,1,2,3,… into two-digit slots breaks down once the numbers reach 100 and no longer fit, and the whole answer depends on where the resulting carries land. So the guess is turned into an exact claim and checked by algebra: a block B of length n works precisely when 9801 · B=10ⁿ-1, and that single equation can be verified by one shift-and-subtract, with no infinite series and no hand-waved carrying. Finally the digit total itself is what proves the block cannot be shortened, so the period question is answered rather than assumed.

1STEP 1

Turn the block into one equation

A repeating block is one clean equation.

0.B=B/(10ⁿ-1) ⟹ 1/9801=B/(10ⁿ-1) ⟺ 9801 B=10ⁿ-1
2STEP 2

Test the one-digit version first

A simpler version suggests the pattern.

1/81=0.012345679 ⟹ guess 1/9801=0.000102…969799
3STEP 3

Multiply the guess by 99

Multiplying back collapses the guess.

100N-N: 00-99_borrow→ 1, 99-97-1=1, (99-k)-(98-k)=1 ⟹ 99N=Σ_j=0⁹⁸100^j
4STEP 4

Multiply by 99 once more

A second multiplication proves it exactly.

99R=100R-R=100⁹⁹-1 ⟹ 9801N=100⁹⁹-1=10¹⁹⁸-1
5STEP 5

Add the digits of the block

The digits add to 883.

2 · 10·(0+1+…+9)=2 · 10 · 45=900, 900-(9+8)=883
6STEP 6

Show 198 is the shortest period

A primality argument shows the period is shortest.

d ∣ 198, 883=198/d · s, 883 prime ⟹ 198/d=1 ⟹ n=198, s=883 → (B)
Answer
883
Two independent checks. First, the period: 9801=3⁴ · 11², and the smallest power of 10 that leaves remainder 1 is 10⁹ for 81 and 10²² for 121, so for 9801 it is 10^lcm(9,22)=10¹⁹⁸. That confirms n=198 by a route with no digit counting in it. Second, the answer choices: the block is the full list 00 through 99 minus one chunk, so the total must be 900 minus a digit total between 0 and 18, i.e. between 882 and 900. That alone kills (A) 874. Among the survivors, 900-883=17 needs the missing chunk to be 89 or 98, while 887, 891, 892 would need chunks with digit totals 13, 9, 8; only the counting order, which ends … 96,97,99, decides between them, and the exact identity 9801N=10¹⁹⁸-1 settles it as 98. Finally the one-digit rehearsal agrees: 1/81=0.012345679 has digit total 45-8=37, the same 'all of them minus the skipped one' shape.
💡Key takeaway

1/99² just counts by twos of digits — 00,01,02,…,97, then jumps to 99 — so its digits add to all of 00 through 99 (900) minus the skipped 98, which is 883.

  • Turn the block into one equation
  • Test the one-digit version first
  • Multiply the guess by 99
  • Multiply by 99 once more
  • Add the digits of the block
  • Show 198 is the shortest period