2.7 Divisibility by 8

Rule: The number formed by the last three digits is divisible by 8. Reasoning: 1000 is divisible by 8 (1000 = 8×125), so everything except the last 3 digits is automatically a multiple of 8.

Example 1: 4,57,816 → last three digits = 816. 816 ÷ 8 = 102 → divisible by 8. Example 2: 34,562 → last three digits = 562. 562 ÷ 8 = 70.25 → not divisible by 8. Example 3: 74,832 → last three digits = 832. 832 ÷ 8 = 104 exact → divisible by 8. Notice you never had to touch the “74” part of the number at all — this is what makes the last-3-digit family of rules (for 8, and for 16) so fast under exam time pressure.

Example 4 (edge case — “000” ending): Is 5,00,000 divisible by 8? Last three digits = “000.” Since 0÷8=0 remainder 0, this counts as divisible. 5,00,000 is divisible by 8 (500000÷8=62500 exact ✓) — the same “zero-ending” trap as in the divisibility-by-4 rule, now for three trailing zeros.

Example 5 (edge case — multiple valid digits): Find all digits x so that the last three digits “2x8” (i.e., 200+10x+8) are divisible by 8. 200+10x+8 = 208+10x. Testing x=0,…,9: 208 (x=0), 248 (x=4), and 288 (x=8) are each divisible by 8 (208÷8=26, 248÷8=31, 288÷8=36, all exact). So x 0,4,8 — three valid digits, again showing that a single missing-digit condition does not always pin down a unique answer; the tens-place coefficient (10) shares a common factor with the modulus (8), which is exactly why multiple digits satisfy it.