Why do some rational numbers have repeating decimal representations? Imagine calculating 17 using long division. You are dividing by 7. What are the possible remainders at each step? They can only be 1, 2, 3, 4, 5, or 6 (why not 0?).

Step-by-step solution

Idea: In long division each step depends only on the current remainder. If a remainder ever comes back, the whole pattern repeats. With only 6 possible remainders, one must come back.

1. Dividing by 7, every remainder is one of 0, 1, 2, 3, 4, 5, 6.½ mark
2. Why not 0? Getting remainder 0 after k steps would mean 7 divides 1 followed by k zeros, i.e. 10^k = 2^k × 5^k. But 7 is a prime that is not 2 or 5, so it never divides 10^k. So the remainder is never 0.1 mark
3. For 17: 10 ÷ 7 = 1 r 3; 30 ÷ 7 = 4 r 2; 20 ÷ 7 = 2 r 6; 60 ÷ 7 = 8 r 4; 40 ÷ 7 = 5 r 5; 50 ÷ 7 = 7 r 1. Remainder 1 is back, which is where we started.1 mark
4. From here the same steps repeat, so the digits 142857 repeat for ever: 17 = 0.142857. With only 6 possible remainders, a repeat must come within 6 steps, so the repeating block has at most 6 digits.½ mark

Check: 0.142857 × 7 = 0.999999, and 142857 × 7 = 999999 = 10⁶ − 1 ✓.

Answer to write in the exam

Common mistakes that cost marks

• Saying the remainders can be 1 to 7. A remainder is always less than the divisor.
• Saying ‘remainder 0 is impossible because 1 is less than 7’. The real reason is that 7 does not divide any power of 10.
• Thinking the repeating block for 17 could be longer than 6 digits.

How this can come in the exam

Try one yourself

Divide 1 by 7 and 3 by 7. How are the two repeating blocks related?

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