Perform the long division for 113. Identify the repeating block of digits. Does it show cyclic properties if you evaluate 213? Now compute 313, 413, etc. What do you notice?

Step-by-step solution

Idea: For 17 all six remainders 1–6 appear in one loop, so every k7 is a rotation of 142857. For 13 there are 12 possible remainders but the loop for 113 only uses 6 of them, so the other 6 form a second loop.

1. Long division of 1 by 13: 10 ÷ 13 = 0 r 10; 100 ÷ 13 = 7 r 9; 90 ÷ 13 = 6 r 12; 120 ÷ 13 = 9 r 3; 30 ÷ 13 = 2 r 4; 40 ÷ 13 = 3 r 1. Remainder 1 is back, so the digits repeat.1 mark
2. 113 = 0.076923076923… = 0.076923; the repeating block is 076923 (6 digits). The remainders used were 10, 9, 12, 3, 4, 1.½ mark
3. 213: 20 ÷ 13 = 1 r 7; 70 ÷ 13 = 5 r 5; 50 ÷ 13 = 3 r 11; 110 ÷ 13 = 8 r 6; 60 ÷ 13 = 4 r 8; 80 ÷ 13 = 6 r 2 → 0.153846. The digits 153846 are different from 076923, so doubling does not just rotate the block: 076923 is not a full cyclic number.1 mark
4. 313 = 0.230769 and 413 = 0.307692: both are rotations of 076923. Continuing, 913, 1013, 1213 also give rotations of 076923, while 513, 613, 713, 813, 1113 give rotations of 153846.1 mark
5. What we notice: every k13 has a 6-digit repeating block, and the blocks come in two cycles of six. 13 has 12 possible remainders but each loop uses only 6, unlike 7 where one loop uses all 6 remainders.½ mark

Check: 076923 × 13 = 999999 and 153846 × 13 = 1999998 = 2 × 999999 ✓.

Answer to write in the exam

Common mistakes that cost marks

• Dropping the leading 0 and writing 113 = 0.76923. The first digit after the point is 0 (10 ÷ 13 = 0).
• Assuming 13 behaves like 7 and claiming all multiples are rotations without checking 213.
• Stopping the division before the remainder repeats and guessing the block.

How this can come in the exam

Try one yourself

Using 213 = 0.153846, find 513.

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