9. Word Problems Combining Divisibility with Number System

These formats appear frequently in SSC CGL Tier-II and RRB NTPC, combining divisibility with LCM/remainder concepts from the broader Number System chapter.

Example 1: Find the largest 4-digit number exactly divisible by 7, 11, and 13. Since 7, 11, 13 are pairwise co-prime, the number must be divisible by their LCM =7×11×13=1001. Multiples of 1001: 1001×1=1001, ×2=2002, …, ×9=9009, ×10=10010 (5-digit, too big). Largest 4-digit multiple = 9009.

Example 2: How many numbers from 1 to 500 are divisible by 3 or 5? Using the inclusion-exclusion principle: (numbers divisible by 3) + (numbers divisible by 5) − (numbers divisible by both, i.e., by 15). ⌊500/3⌋=166, ⌊500/5⌋=100, ⌊500/15⌋=33. Total =166+100-33=233.

Example 3: Find the smallest number which, when divided by 6, 7, or 8, always leaves a remainder of 3. The number must be 3 more than a common multiple of 6, 7, and 8. LCM(6,7,8)=168 (since 8=2^3 is the highest power of 2 involved, 7 is prime, 3 from 6 — LCM =168). Smallest number =168+3=171. Check: 171÷6=28 remainder 3; 171÷7=24 remainder 3; 171÷8=21 remainder 3. ✓

Example 4: A 3-digit number is formed using digits 4, 7, and 2 (each used once) such that the number is divisible by 4. How many such arrangements are possible? Arrangements of {4,7,2}: 472, 427, 742, 724, 247, 274 (6 total). Test last-two-digit rule for 4: 72÷4=18✓(472); 27÷4 no(427); 42÷4 no(742); 24÷4=6✓(724); 47÷4 no(247); 74÷4 no(274). 2 arrangements (472 and 724) are divisible by 4.

Example 5: What is the remainder when 200! is divided by 202? (Trick question testing conceptual understanding, not brute computation.) 202 = 2×101, and 101 is a prime number greater than the numbers being multiplied in 200! only if 101 > 200 — but here 101 ≤ 200, so 101 is one of the factors inside 200! (since 200! = 1×2×…×101×…×200). Hence 200! is divisible by 101, and clearly by 2 as well (being a huge factorial), so 200! is divisible by 202. Remainder = 0. (Contrast this with the classic Wilson’s-theorem-style question “find remainder of (p-1)! divided by p for prime p,” which is a more advanced variant occasionally seen in CGL Tier-II — for prime p, (p-1)! ≡ -1 (mod p), i.e., remainder =p-1.)

Example 6 (harder — combining LCM with a common remainder): Find the smallest number which, when divided by 9, 12, or 15, always leaves a remainder of 5. The number must be 5 more than a common multiple of 9, 12, and 15. 9=3^2, 12=2^2×3, 15=3×5, so LCM(9,12,15)=2^2×32×5=180. Smallest number =180+5=185. Check: 185÷9=20 remainder 5; 185÷12=15 remainder 5; 185÷15=12 remainder 5. ✓

Example 7 (largest number instead of smallest): Find the largest 5-digit number that is exactly divisible by 12, 15, and 18. 12=2^2×3, 15=3×5, 18=2×3^2, so LCM(12,15,18)=2^2×32×5=180. The largest 5-digit number is 99,999. 99999 ÷ 180 = 555 remainder 99 (since 180×555=99900). Largest 5-digit multiple of 180 =99900.