Number System

Find the LCM of 10, 12, 15, 20. LCM(10, 12, 15, 20) = 60. The number is of the form 60k + 4. For k=1, the number is 64.

The LCM of 15, 20, 28 is 420. The greatest 4-digit number is 9999. Dividing 9999 by 420 gives a quotient of 23 and remainder 339. The number x is 420 * 23 + 2 = 9662. The sum of digits is 9 + 6 + 6 + 2 = 23.

The number must divide (772 - 5) = 767 and (2778 - 5) = 2773. We find the greatest common divisor of 767 and 2773. 767 = 13 * 59. 2773 / 59 = 47. Since 59 is a common factor, the greatest value is 59.

Product of two numbers = HCF * LCM. 26010 = HCF * 510. HCF = 26010 / 510 = 51.

LCM(12, 16, 54) = 432. Number = 432k + 7. We need 432k + 7 to be divisible by 13. 432 mod 13 = 3. So (3k + 7) mod 13 = 0. For k=2, 6+7=13 (divisible). Number = 432(2) + 7 = 864 + 7 = 871. Sum of digits = 8+7+1 = 16.

The number must be of the form LCM(15, 18, 24) * k + 8. The LCM is 360, so the number is 360k + 8. We need 360k + 8 to be divisible by 13; testing values of k, 360k + 8 = 27k + 8 (mod 13), which simplifies to k + 8 = 0 (mod 13), so k = 5. The number is 360(5) + 8 = 1808, and the sum of digits is 1+8+0+8 = 17.

Let the number be N and the sum of digits be S=14. We are given N = 51 * 14 + 11, which equals 714 + 11 = 725. The sum of digits of 725 is 7 + 2 + 5 = 14, which matches the condition.

If 42705 and 43960 leave the same remainder when divided by n, then n must divide their difference: 43960 - 42705 = 1255. Factors of 1255 are 1, 5, 251, 1255. The only 3-digit factor is 251.

The correct answer is C. If two numbers leave the same remainder when divided by a certain number, then their difference is divisible by that number. Here, 34041 - 32506 = 1535. We need a 3-digit divisor of 1535. 1535 = 5 × 307. The 3-digit divisors are 307 and 5 × 307 = 1535 (not 3-digit). So the only 3-digit divisor is 307.

The number must be 3 more than a multiple of LCM(5,6,7,8) = 840. So it's of form 840k + 3, and also divisible by 9. Testing: k=1 gives 843 (sum=15, not divisible by 9). k=2 gives 1683 (sum=18, divisible by 9). k=3 gives 2523 (sum=12, not divisible by 9). The LCM condition requires the remainder 3 for each divisor, so 1683 is the least such number. Option A (1793) gives wrong remainders.

Let N be the number. N = 48q + 21 for some integer q. Since 16 is a factor of 48, we can write 48q + 21 = 16(3q) + 16 + 5 = 16(3q + 1) + 5. Therefore, when N is divided by 16, the remainder is 5.

For any two numbers: Product = LCM × HCF. Given LCM = 4125, HCF = 25, and first number = 375. Second number = (LCM × HCF) ÷ First number = (4125 × 25) ÷ 375 = 275. Difference = 375 - 275 = 100. Therefore, the second number is 100 less than the first. Options A, B, C (25, 50, 75) are incorrect as they don't satisfy the LCM-HCF product relationship.

We need the greatest number that divides 690 and 875 leaving remainders 10 and 25. This means the number divides (690-10) = 680 and (875-25) = 850 exactly. Find HCF(680, 850). 680 = 2³ × 5 × 17, 850 = 2 × 5² × 17. HCF = 2 × 5 × 17 = 170. Option B (130) is a common divisor but not the greatest; options C, D are not factors of both adjusted numbers.