2.12 Divisibility by 13

Rule (Osculator method, additive): Multiply the last digit by 4 and add it to the remaining number. Repeat; if the result is divisible by 13, so is the original. Reasoning: The positive osculator for 13 is 4, because 10 ≡ -3 (mod 13), and multiplying by -13·(something)… practically: if N=10a+b, then N ≡ 0 (mod 13) 10a+b≡0 -3a+b≡0 3a≡ b a ≡ 9b ≡ -4b (mod 13) — equivalently a+4b ≡ 0 (mod 13) (since 9≡ -4, sign works out to give the add-4b rule commonly taught).

Example 1: Check 2,731. Last digit 1, ×4 = 4. Remaining = 273. 273+4 = 277. Again: last digit 7, ×4=28. Remaining=27. 27+28=55. 55 ÷ 13 = 4.23 → not divisible. (Check: 2731 ÷ 13 = 210.08 → correctly not divisible.)

Example 2: Check 1,001. Last digit 1, ×4=4. Remaining=100. 100+4=104. 104 ÷ 13 = 8 exactly → divisible by 13. (1001 = 7×11×13 ✓)

Example 3 (1001-grouping trick): Check 6,214,357 for divisibility by 13 using groups of 3 from the right: 357, 214, 6. Alternating sum = 357-214+6=149. 149 ÷ 13 = 11.46 → not divisible by 13.

Example 4: Check whether 5,876 is divisible by 13. Last digit 6, ×4=24. Remaining=587. 587+24=611. Again: last digit 1, ×4=4. Remaining=61. 61+4=65. 65÷13=5 exact → 5,876 is divisible by 13 (direct check: 5876÷13=452 exact ✓).

Example 5: Check whether 10,921 is divisible by 13, and find the remainder if not. Last digit 1, ×4=4. Remaining=1092. 1092+4=1096. Again: last digit 6, ×4=24. Remaining=109. 109+24=133. Again: last digit 3, ×4=12. Remaining=13. 13+12=25. 25÷13=1 remainder 12 — not exactly divisible. Direct check confirms 10921÷13=840 remainder 1 (the small residual number from repeated osculation carries the same divisibility status as the original, though its raw remainder value differs from the original’s remainder — the method only guarantees agreement on whether it’s divisible by 13, not on the exact remainder). 10,921 is not divisible by 13; its true remainder on division by 13 is 1.