2.20 Divisibility Rules Beyond 19 (Composite Divisors Frequently Asked)

Figure: For a composite divisor, split it into co-prime factors and check both.

SSC and RRB papers occasionally test divisors above 19. These are all handled by breaking the divisor into co-prime factors and applying rules you already know — there is no need to memorize new osculators for these.

Divisibility by 20: Divisible by both 4 and 5. Equivalently, the last two digits form a number divisible by 20 (i.e., last digit is 0 and the tens digit is even, or last two digits are one of 00, 20, 40, 60, 80). Example: 3,480 → last two digits 80, divisible by 20 → divisible by 20.

Divisibility by 22: Divisible by both 2 and 11. Example: 2,42 → even; alternating sum 2-4+2=0 → divisible by 11 → divisible by 22.

Divisibility by 24: Divisible by both 8 and 3. Example: 1,368 → last three digits 368, 368÷8=46 exact ✓; digit sum 1+3+6+8=18, divisible by 3 ✓ → divisible by 24.

Divisibility by 33: Divisible by both 3 and 11.

Divisibility by 36: Divisible by both 4 and 9.

Divisibility by 40: Divisible by both 8 and 5.

Divisibility by 44: Divisible by both 4 and 11.

Divisibility by 45: Divisible by both 9 and 5.

Divisibility by 99: Since 99 = 9×11, check both 9 (digit sum) and 11 (alternating sum). A useful direct test also exists: group the number into pairs of 2 digits from the right and sum all the pairs; if that sum is divisible by 99, so is the original (parallel to the 1001-grouping trick for 7/11/13, since 10^2≡1(mod 99)). Example: Check 47,52 for divisibility by 99: pairs from right = 52, 47. Sum =52+47=99 → divisible by 99. (Check: 4752 ÷ 99 = 48 exact ✓.)

General principle: For any composite divisor d = p × q where p,q are co-prime, “divisible by d” ⟺ “divisible by p AND divisible by q.” This single principle covers every composite divisor an exam can throw at you — you never need to memorize a fresh rule for 24, 36, 40, 44, 45, 63, 72, 88, 99, etc.; just factor it into co-primes you already have rules for.

Example (divisibility by 45): Is 8,730 divisible by 45? 45=9×5. Last digit 0 → divisible by 5 ✓. Digit sum =8+7+3+0=18, divisible by 9 ✓. Divisible by 45 (8730÷45=194 exact ✓).

Example (divisibility by 88): Is 2,464 divisible by 88? 88=8×11. Last three digits = 464, 464÷8=58 exact ✓. Alternating sum (digits from right: 4,6,4,2): odd positions 4+4=8, even positions 6+2=8, difference =0 → divisible by 11 ✓. Divisible by 88 (2464÷88=28 exact ✓).