Patterns & Reasoning

Problem

Some decimals repeat forever

A division gives a decimal that never ends. We need two digits from far down in it.
Number system
Your answer
How to solve
Strategy Look for a Pattern — Twenty places of long division is not the intended work. Find the block that repeats and its length, then ask where each place lands inside a block -- a remainder question, done twice.
1STEP 1

Clear the decimal from the divisor

Multiplying both numbers by 10 leaves the quotient unchanged and makes the division easier to see.

14 ÷ 2.7 = 140 ÷ 27
2STEP 2

Divide until it repeats

The digits 1, 8, 5 come out and then the same remainder returns, so the pattern starts over.

140 ÷ 27 = 5.185185…
3STEP 3

Locate the 15th place

Divide 15 by the block length of 3. It comes out exactly, so the 15th place is the last digit of a block.

15 ÷ 3 = 5
4STEP 4

Locate the 20th place

This time there is a remainder of 2, so the 20th place is the second digit of a block.

20 = 3 × 6 + 2
Answer
15th place: 5, 20th place: 8
Count out the first few: places 1 to 6 are 1, 8, 5, 1, 8, 5, so every place that is a multiple of 3 holds 5, and every place two past a multiple holds 8.
Takeaway

A far-off digit of a repeating decimal is decided by a remainder, not by finishing the division.

  • Clear the decimal from the divisor
  • Divide until it repeats
  • Locate the 15th place
  • Locate the 20th place

▶ Practice — 12 problems