2.6 Divisibility by 7

Rule (Osculator method): Double the last digit, subtract it from the remaining number (the number formed by all digits except the last). Repeat until you get a small number. If that number is 0 or divisible by 7, the original is divisible by 7. Reasoning: This works because 10 ≡ 3 (mod 7), and the “double-and-subtract” operation is equivalent to multiplying the truncated part by a factor that preserves divisibility by 7 (the “negative osculator” for 7 is -2, i.e., double-and-subtract). A full modular proof: if N = 10a + b (where b is the last digit), then N is divisible by 7 iff a - 2b is divisible by 7, because 10a+b ≡ 0 (mod 7) 3a+b≡0 multiplying by -2 (invertible mod 7): -6a - 2b ≡ 0 a - 2b ≡ 0 (mod 7) (using -6≡1).

Example 1: Check 2016. Last digit 6, double = 12. Remaining number = 201. 201 - 12 = 189. Repeat: last digit 9, double = 18. Remaining = 18. 18-18=0 → divisible by 7. (Check: 2016 = 7×288 ✓)

Example 2: Check 5,432. Last digit 2, double = 4. Remaining = 543. 543-4=539. Again: last digit 9, double=18. Remaining=53. 53-18=35, which is divisible by 7 → 5,432 is divisible by 7. (Check: 5432 = 7 × 776 ✓)

Example 3 (competitive shortcut using 1001 = 7×11×13): Since 1001 = 7 × 11 × 13, for a large number, split it into groups of 3 digits from the right and take the alternating sum. That alternating sum’s divisibility by 7 (or 11 or 13) tells you about the original number’s divisibility by that prime. Check 4,753,296 by 7: groups from right → 296, 753, 4. Alternating sum = 296 - 753 + 4 = -453. Now check if -453 (i.e., 453) is divisible by 7: 453 ÷ 7 = 64.71 → not divisible → original number not divisible by 7.

Example 4 (a case that fails): Check whether 9,847 is divisible by 7. Last digit 7, double = 14. Remaining = 984. 984-14=970. Again: last digit 0, double = 0. Remaining = 97. 97-0=97. Again: last digit 7, double=14. Remaining=9. 9-14=-5. Since -5 (i.e. 5) is not 0 or a multiple of 7, 9,847 is NOT divisible by 7 (direct check: 9847÷7=1406.71, remainder 5 — consistent).

Example 5 (missing-digit type): Find the digit x so that the 3-digit number 62x (i.e., 620 + x) is divisible by 7. Testing x=0 to 9 directly is fastest here since the number is small: 623 = 7×89 works exactly. x=3 is the unique digit (check: 620+3=623, and 623÷7=89 exact ✓; no other digit from 0–9 gives a multiple of 7 in this range).