Number System

Subtract remainder from both numbers: 245-5 = 240 and 1029-5 = 1024. Find GCD of 240 and 1024. 240 = 16 × 15, 1024 = 16 × 64. GCD is 16. Verification: 240/16 = 15 remainder 0, 1024/16 = 64 remainder 0. 12 doesn't divide 1024, 14 doesn't divide 240 evenly, 20 is not the GCD.

When a number N = 345k + 69 is divided by 15, the term 345k is always divisible by 15 since 345 = 15 × 23. Therefore, the remainder depends only on 69 divided by 15, which gives 69 = 15 × 4 + 9. So the remainder is 9.

When a number leaves remainder 75 when divided by 2736, the number can be written as 2736k + 75. Since 2736 = 24 × 114 is exactly divisible by 24, we only need to find 75 mod 24. 75 = 3 × 24 + 3, so the remainder is 3.

If 2270, 3739, and 6677 all give same remainder r when divided by d, then: (3739-2270) = 1469, (6677-3739) = 2938, (6677-2270) = 4407. All differences must be divisible by d. Finding GCD: 2938 = 2 × 1469, 4407 = 3 × 1469. So d must divide 1469. 1469 = 13 × 113. Checking options: 113 is a three-digit divisor and 1469 is divisible by it.

Number giving remainder 5 when divided by 8, 10, 12 is LCM(8,10,12) × k + 5 = 120k + 5. For divisibility by 13: 120k + 5 ≡ 0 (mod 13), so 120k ≡ 8 (mod 13). 120 ≡ 3 (mod 13), so 3k ≡ 8 (mod 13). Testing k = 7: 3×7 = 21 ≡ 8 (mod 13). Number = 120×7 + 5 = 845. Check: 845 ÷ 13 = 65 exactly.

Working backwards from the successive divisions: if a number leaves remainder 4 when divided by 8, that quotient leaves remainder 5 when divided by 7, and that quotient leaves remainder 3 when divided by 6, the least such number is 201. When 201 is divided successively by 8, 7, 6: 201 ÷ 8 = 25 remainder 1, 25 ÷ 7 = 3 remainder 4, 3 ÷ 6 = 0 remainder 3. The final remainder is 3.

To find the largest number that divides 61, 72, and 83 leaving remainders 1, 2, and 3 respectively, we need to find the HCF of (61-1)=60, (72-2)=70, and (83-3)=80. HCF of 60, 70, and 80 is 10. Options A (20), B (15), and C (12) do not divide all three numbers evenly.

If remainders are equal, the differences between numbers must be divisible by the divisor. Differences: 151-76=75, 226-151=75, 226-76=150. The HCF of these differences is 75. Checking: 76÷75 leaves remainder 1, 151÷75 leaves remainder 1, 226÷75 leaves remainder 1. All remainders equal.

If remainders are equal, the divisor divides differences: 6677-3739=2938, 3739-2270=1469, 6677-2270=4407. GCD(2938,1469,4407)=1469. So divisor is a 3-digit factor of 1469. 1469=17×7×13. Checking options: only 113 gives remainder 6 with all three numbers.

Using successive division: N = 4a + 2, a = 5b + 3, b = 6c + 4. Working backwards: b = 6c + 4, a = 5(6c + 4) + 3 = 30c + 23, N = 4(30c + 23) + 2 = 120c + 94. When N is divided by 60, N = 60(2c) + 34, remainder = 34. Alternatively, use Chinese remainder theorem or direct substitution.

The number can be written as 7560k + 85. Since 7560 is divisible by 24 (7560 = 24 × 315), we only need the remainder of 85 divided by 24. 85 = 24 × 3 + 13, so the remainder is 13. Option B is correct.

When a number d leaves the same remainder when dividing a, b, and c, then d divides the differences: (a-b), (b-c), (a-c). Find GCD of these differences. Here: |392-486|=94, |486-627|=141, |392-627|=235. GCD(94,141,235)=47 since 94=2×47, 141=3×47, 235=5×47.

Find LCM of 12, 15, 20, 54. Prime factors: 12=2²×3, 15=3×5, 20=2²×5, 54=2×3³. LCM = 2²×3³×5 = 540. The number must be 540k + 4. Smallest such number is 540 + 4 = 544. Verify: 544 ÷ 12 = 45R4, 544 ÷ 15 = 36R4, 544 ÷ 20 = 27R4, 544 ÷ 54 = 10R4.

HCF = 15, LCM = 20 × 15 = 300. Quotient of LCM/HCF = 300/15 = 20. First number = 4 × (20 - 5) = 4 × 15 = 60. For any two numbers, Product = HCF × LCM. So 60 × other = 15 × 300 = 4500. Therefore other = 4500/60 = 75.

Number leaves remainder 3 when divided by 5, 6, or 8, so it equals LCM(5,6,8)×k + 3 = 120k + 3. It must also be divisible by 9. Testing values: 123 (k=1) not divisible by 9; 243 (k=2) = 243, which is divisible by 9 (2+4+3=9). Verify: 243 ÷ 5 = 48 rem 3, 243 ÷ 6 = 40 rem 3, 243 ÷ 8 = 30 rem 3, 243 ÷ 9 = 27 rem 0.