Number System

If numbers leave the same remainder when divided by x, then the differences between the numbers are divisible by x. Differences: 2633-2460=173 and 2806-2633=173. Since 173 is prime, x can be 173 or its factors. Sum of digits of 173 is 1+7+3=11.

For divisibility by 16, only the last 4 digits matter. 81 ÷ 16 = 5 remainder 1.

If HCF of two numbers is 12, then both numbers are multiples of 12. Let the numbers be 12a and 12b where a and b are coprime. Their LCM = 12ab must be a multiple of 12. Checking the options: 72 = 12×6 (multiple of 12, possible), 60 = 12×5 (multiple of 12, possible), 84 = 12×7 (multiple of 12, possible), but 90 = 12×7.5 (not an integer multiple of 12, so 90 cannot be the LCM). The LCM must always be divisible by the HCF.

Finding HCF of 2040 and 391. 391 = 17 × 23 (since 391/17 = 23). Checking if 17 divides 2040: 2040/17 = 120, which is an integer. Checking if 23 divides 2040: 2040/23 = 88.7, not an integer. Therefore HCF = 17.

Numbers are in ratio 2:3:5, so let them be 2k, 3k, 5k. Their LCM = 90. 2k, 3k, 5k have LCM = 30k (since 2, 3, 5 are coprime). So 30k = 90, giving k = 3. The numbers are 6, 9, 15. Their HCF = 3. Alternatively: LCM = k × LCM(2,3,5) = k × 30 = 90, so k = 3. HCF = k × HCF(2,3,5) = 3 × 1 = 3.

For two numbers, HCF × LCM = product of the numbers. Given HCF = 6 and LCM = 120, the product is 720. Checking option A: 24 × 30 = 720 ✓ and HCF(24,30) = 6 ✓ and LCM(24,30) = 120 ✓. Option B: 18 × 40 = 720 but HCF = 2, not 6. Option C: 12 × 60 = 720 but HCF = 12, not 6.

The smallest 4-digit number is 1000. Find LCM of 24 and 36: 24 = 2³ × 3, 36 = 2² × 3². LCM = 2³ × 3² = 8 × 9 = 72. Find the smallest 4-digit number divisible by 72: 1000 ÷ 72 = 13.89..., so we need 14 × 72 = 1008. The remainder when 1008 is divided by 40: 1008 ÷ 40 = 25.2, so 25 × 40 = 1000, remainder = 1008 - 1000 = 8. The square of 8 = 64.

Find LCM of 60, 18, 20. 60 = 2² × 3 × 5. 18 = 2 × 3². 20 = 2² × 5. LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180. So the number N = 180k + 3. For the largest 4-digit number: 180k + 3 ≤ 9999. 180k ≤ 9996. k ≤ 55.5, so k = 55. N = 180 × 55 + 3 = 9900 + 3 = 9903. This is a 4-digit number that leaves remainder 3 when divided by 60, 18, and 20. Let's verify: 9903 ÷ 60 = 165 remainder 3. 9903 ÷ 18 = 550 remainder 3. 9903 ÷ 20 = 495 remainder 3. The answer is indeed 9903.

For the same remainder r when dividing 49, 147, 322 by d: 49 = dq1 + r, 147 = dq2 + r, 322 = dq3 + r. Subtracting pairs eliminates r: differences are 98, 175, 253. The HCF of these differences gives the greatest possible d. HCF(98, 175, 253) = 7. Verification: 49 ÷ 7 = 7 rem 0, 147 ÷ 7 = 21 rem 0, 322 ÷ 7 = 46 rem 0. All give remainder 0, confirming 7 is the answer.

363 = 33 × 11, so any remainder mod 363 gives the same remainder mod 11. 17 ÷ 11 leaves remainder 6.

The number must be 3 more than a multiple of LCM(7,9,11) = 693. 99795 - 3 = 99792, which is divisible by 693 (144 × 693), and 99795 is the largest 5-digit number satisfying this.

Statement I: Number gives remainder 2 when divided by 5, so it could be 12, 22, 32, 42, 52, 62, 72, 82, 92. Some of these are divisible by 12 (12, 24, 36, 48, 60, 72, 84, 96), some are not. Statement II: Number gives remainder 1 when divided by 3. Combining both statements: numbers with remainder 2 mod 5 are 12, 22, 32, 42, 52, 62, 72, 82, 92. Checking which give remainder 1 mod 3: only 22, 52, 82 satisfy both. None of these are divisible by 12. But we cannot conclude divisibility without checking each candidate - and there's no unique determination. Even together, insufficient.

7^2 = 49 = 2*25 - 1. So 7^2 is congruent to -1 (mod 25). Then 7^2015 = 7 * (7^2)^1007 is congruent to 7 * (-1)^1007 = -7 (mod 25). -7 is congruent to 18 (mod 25).

Numbers 1-25. Composite: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25 (15 total). Non-composite: 1, 2, 3, 5, 7, 11, 13, 17, 19, 23 (10 total). Composite divisible by 5: 10, 15, 20, 25 (4). Non-composite divisible by 2: 2 (1). Total favorable = 5. Probability = 5/25 = 0.2.

We need 2^403 mod 15. Note 2^4 = 16 = 1 mod 15. So 2^403 = 2^(4*100 + 3) = (2^4)^100 * 2^3 = 1^100 * 8 = 8 mod 15. Thus k = 8.