8. Divisibility in Consecutive Integers

Figure: Among any n consecutive integers, exactly one is always divisible by n.

Key facts (very frequently tested): 1. The product of any 2 consecutive integers is always divisible by 2. 2. The product of any 3 consecutive integers is always divisible by 6 (since among any 3 consecutive numbers, at least one is divisible by 2 and exactly one is divisible by 3, and 2×3=6). 3. The product of any 4 consecutive integers is always divisible by 24. 4. More generally, the product of any n consecutive integers is divisible by n!. 5. Among any n consecutive integers, exactly one is divisible by n. 6. The product of any 3 consecutive integers, when one of them is even and increased appropriately, connects to n(n+1)(n+2) being divisible by 6 always — useful for “prove divisibility” reasoning questions.

Example 1: Show that n(n+1)(n+2) is always divisible by 6 for any positive integer n. Among 3 consecutive integers, at least one is even (divisible by 2) and exactly one is divisible by 3 (since every 3rd integer is a multiple of 3). Since 2 and 3 are co-prime, the product is divisible by 2×3=6.

Example 2: Is 17×18×19×20 divisible by 24? Yes — it’s a product of 4 consecutive integers, always divisible by 4!=24. Verify: 17×18×19×20 = 116280. 116280 ÷ 24 = 4845 exactly. ✓

Example 3: n^3-n is always divisible by which number? n^3-n = n(n-1)(n+1), the product of 3 consecutive integers — always divisible by 6.

Example 4: Is the product of any 5 consecutive positive integers always divisible by 120? Yes — by the general rule, the product of n consecutive integers is always divisible by n!, and 5!=120. Verify with 6×7×8×9×10 = 30240; 30240 ÷ 120 = 252 exact. ✓

Example 5 (a general algebraic identity, harder): Show that n^5-n is always divisible by 30 for any positive integer n. n^5-n = n(n^4-1) = n(n^2-1)(n2+1) = n(n-1)(n+1)(n^2+1). The factor n(n-1)(n+1) is a product of 3 consecutive integers, so it is always divisible by 6 (Example 1). A separate number-theoretic check shows n^5-n is additionally always divisible by 5 (by Fermat’s Little Theorem, n^5≡ n5 for every integer n). Since (6,5)=1, n^5-n is divisible by 6×5=30. Verify for n=2: 2^5-2=30, and 30÷30=1 exact. For n=7: 7^5-7=16807-7=16800, and 16800÷30=560 exact. ✓

Example 6 (edge case — a stronger divisor is NOT guaranteed): Is the product of any 5 consecutive positive integers always divisible by 720 (i.e., 6!)? Take the concrete case 23×24×25×26×27. By the general rule this product is guaranteed divisible by 5!=120 — verify: the product equals 9,686,520, and 9686520÷120=80721 exactly. But checking against 720: 9686520÷720=13453.5, not exact. So the product of 5 consecutive integers is guaranteed divisible by 120 but not guaranteed divisible by 720 — a common trap where aspirants over-generalize the “n consecutive integers divisible by n!” rule and assume a larger factorial also always divides, when in fact only n! itself is guaranteed.

Example 4 (negative remainder trick): Find the remainder when 59 is divided by 7. Direct: 59 = 7×8+3, remainder 3. Alternatively, note 59 = 63-4 = 7×9-4, so the remainder can be seen as -4, which converts to a positive remainder as 7-4=3. This “negative remainder” trick is extremely useful when a number is just below a convenient multiple — e.g., finding the remainder of 6398 divided by 100 is trivially 98, but finding the remainder of 6398 divided by 99 is faster via 6398 = 6400-2 = 64×100-2, and since 100≡1(mod 99), 6398≡64-2=62(mod 99).

Example 5 (combining two remainder conditions): A number leaves remainder 3 when divided by 5, and remainder 4 when divided by 7. Find the smallest such positive number. List numbers leaving remainder 3 on division by 5: 3, 8, 13, 18, 23, … Check each against “remainder 4 on division by 7”: 3÷7 rem 3 (no); 8÷7 rem 1 (no); 13÷7 rem 6 (no); 18÷7 rem 4 (yes!). Smallest number = 18.