Try to find the decimal expansions of 103 and 1112. What do you observe about the repetition of the digits after the decimal point?

Step-by-step solution

Idea: Do both long divisions and track the remainders. Once a remainder repeats, so do the digits. Whether repetition starts at once depends on the denominator: 12 = 2² × 3 has factors of 2 that give a few non-repeating digits first.

1. 103: 10 ÷ 3 = 3 remainder 1; then 10 ÷ 3 = 3 remainder 1 again and again. So 103 = 3.333… = 3.3.1 mark
2. 1112: 110 ÷ 12 = 9 remainder 2; 20 ÷ 12 = 1 remainder 8; 80 ÷ 12 = 6 remainder 8; 80 ÷ 12 = 6 remainder 8 … So 1112 = 0.91666… = 0.916.1 mark
3. Observation: both are non-terminating and repeating, each with a one-digit repeating block. In 103 the repetition starts immediately after the decimal point (pure repeating); in 1112 two digits (9, 1) come first and then 6 repeats (general repeating). The repetition begins as soon as a remainder repeats.1 mark

Check: 3.3 × 3 = 9.999… = 10 ✓; 0.916 × 12 = 11 (0.91666… × 12 = 10.99999…) ✓.

Answer to write in the exam

Common mistakes that cost marks

• Writing 1112 = 0.916. Only the 6 repeats; 9 and 1 appear once.
• Writing 103 = 0.333…, forgetting the whole-number part 3.
• Stopping the division after one or two digits and calling the decimal terminating.

How this can come in the exam

Try one yourself

Find the decimal expansion of 136.

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