Number System

Any number that leaves remainder 29 when divided by 56 can be written as 56k + 29 for some integer k. When this number is divided by 8, we get 8(7k + 3) + 5, so the remainder is 5. This works because 56 itself is a multiple of 8, so only 29 contributes to the remainder.

The greatest 5-digit number is 99999. We need a number that leaves remainder 2 when divided by 3, 5, 8, and 12. This means N-2 must be divisible by the LCM of 3, 5, 8, 12. LCM(3,5,8,12) = 120. The largest multiple of 120 less than 99999 is 99960. Adding 2 gives 99962. Checking: 99962 ÷ 3 = 33320 remainder 2, 99962 ÷ 5 = 19992 remainder 2, etc. Option C (99960) would leave remainder 0, not 2.

When number N is divided by 987, remainder is 59. This means N = 987k + 59 for some integer k. To find remainder when N is divided by 21, we need (987k + 59) mod 21. Since 987 = 21 × 47 (987 is divisible by 21), 987k mod 21 = 0. So N mod 21 = 59 mod 21. Dividing 59 by 21: 21 × 2 = 42, remainder = 59 - 42 = 17. The remainder is 17. Option A (21) is impossible since remainder must be less than divisor.

Test each divisor: 9361 ÷ 11 = 851 exactly with no remainder. For verification: 19×492=9348 (remainder 13), 17×550=9350 (remainder 11), 13×720=9360 (remainder 1). Only 11 divides evenly.

Let the GCD be d. Then 392 = ad + r, 486 = bd + r, 627 = cd + r. Subtracting: 94 = d(b-a), 235 = d(c-b). GCD of 94 and 235 is 47. Checking: 392 ÷ 47 = 8 remainder 16, 486 ÷ 47 = 10 remainder 16, 627 ÷ 47 = 13 remainder 16. The same remainder is 16, so 47 is correct.

HCF is found by prime factorization. 425 = 5² × 17. 476 = 2² × 7 × 17. The only common prime factor is 17, so HCF = 17. Option A (4) is a factor of 476 but not 425. Option B (5) divides 425 but not 476. Option D (51 = 3 × 17) requires 3 as a factor, which neither number has.

For two numbers with HCF 29, let them be 29x and 29y where x,y are coprime. Then LCM = 29xy = 4147, so xy = 143. The pairs are (11,13) giving numbers 319 and 377. Sum = 319+377 = 696.

For two numbers, Product = LCM × HCF. Here: 6188 × 68 = 420,784. Option C: 884 × 476 = 420,784. Both are divisible by 68 (884÷68=13, 476÷68=7, and 13, 7 are coprime). Other pairs don't satisfy both conditions.

Using the relationship: HCF × LCM = product of numbers. 113 × 228825 = 25,857,225. Dividing by 2825 gives 9153. Verify: 113 × 9153 × 2825 = 228825.

When a number N divided by 192 gives remainder 54, we can write N = 192k + 54. Since 192 is exactly divisible by 16 (192 = 16 × 12), the term 192k contributes 0 remainder when divided by 16. Therefore, the final remainder is simply 54 mod 16. 54 = 16 × 3 + 6, so the remainder is 6.

Let the numbers be 33a and 33b where a and b are coprime. Then 33(a+b) = 528, so a+b = 16. Coprime pairs summing to 16 are: (1,15), (3,13), (5,11), (7,9) and their reverses. This gives 4 pairs.

We need N such that N = 12a + 2 = 16b + 6 = 24c + 14. This means N - 10 is divisible by LCM(12, 16, 24) = 48. Largest 4-digit multiple of 48 is 9984. Adding 10 gives 9994, but 9994 - 48 = 9946 doesn't work. Testing backwards: 9974 satisfies all conditions (9974/12 = 831r2, /16 = 623r6, /24 = 415r14).

When n leaves remainder 2 when divided by 5, we can write n = 5k + 2 for some integer k. Then 3n = 3(5k + 2) = 15k + 6 = 5(3k + 1) + 1. This means when 3n is divided by 5, the remainder is 1. Alternatively, simply multiply the remainder: 3 × 2 = 6, and 6 ÷ 5 leaves remainder 1.

LCM of 5,6,7,8 = 840. Number x = 840k + 3. Also x divisible by 9: 840k + 3 ≡ 0 (mod 9). Since 840 ≡ 3 (mod 9), we get 3k + 3 ≡ 0 → k = 2 (minimum). So x = 840(2) + 3 = 1683. Product of digits = 1 × 6 × 8 × 3 = 144. The number must satisfy all congruence conditions simultaneously.

Let n = dq + 23 where 23 < d. Then 2n = d(2q) + 46. The remainder of 46 divided by d is 11. This means 46 = d × 1 + 11, so d = 35. Check: if remainder is 23 and divisor is 35, then 2 × 23 = 46 gives remainder 11 (since 46 - 35 = 11). Options A (36), C (34), and D (20) don't work.