Number System

The product of two numbers equals the product of their HCF and LCM. Therefore, 45 * x = 180 * 15. Solving for x gives x = (180 * 15) / 45 = 60.

The product of two numbers is equal to the product of their HCF and LCM. Given HCF = 25, LCM = 150, and one number = 50, the other number is (25 * 150) / 50 = 75.

If a number N = 899q + 63, then N = (31 * 29)q + (29 * 2 + 5). Since 899 is divisible by 29, the remainder when N is divided by 29 is the same as the remainder when 63 is divided by 29. 63 = 29 * 2 + 5.

Let the number be 10x + y. The reversed number is 10y + x. The difference is |9x - 9y| = 9|x - y|. This must be the HCF. Since the difference is a multiple of 9, the HCF must be 9. Thus, 9|x - y| = 9, so |x - y| = 1. The sum of digits x + y can vary, but in many such problems, the sum is constant. Checking options, 9 is the only digit sum that fits common constraints.

Product of two numbers = LCM * HCF. 15 * x = 1575 * 5. x = (1575 * 5) / 15 = 1575 / 3 = 525.

The LCM of two numbers must be a multiple of their HCF. 85 / 17 = 5, which is an integer. The others (65, 105, 95) are not divisible by 17.

The greatest 3-digit number divisible by 12 is 996 (since 1000/12 is 83.33, 12*83 = 996). The least 4-digit number divisible by 12 is 1008 (since 1000/12 is 83.33, 12*84 = 1008).

The number must divide the differences: (99-82)=17, (150-99)=51, (150-82)=68. The GCD of 17, 51, and 68 is 17.

The HCF of two numbers is a factor of those numbers. If HCF(a,b)=25 and HCF(c,d)=15, the HCF of all four must be a common factor of 25 and 15, which is 5.

Find the LCM of 9, 11, and 13, which is 1287. To leave a remainder of 6, the number must be of the form 1287k + 6. For k=1, the number is 1293. Subtracting 1293 from 1294 gives 1.

To find the greatest number that leaves the same remainder, we find the differences between the numbers: 132 - 62 = 70, 237 - 132 = 105, and 237 - 62 = 175. The HCF of 70, 105, and 175 is 35.

If a number divides 43, 91, and 183 leaving the same remainder, it must divide the differences: 91 - 43 = 48, 183 - 91 = 92, and 183 - 43 = 140. The greatest common divisor of 48, 92, and 140 is 4.

The required number must divide 1625 - 8 = 1617, 2281 - 4 = 2277, and 4218 - 5 = 4213 without leaving any remainder. The greatest common divisor (HCF) of 1617, 2277, and 4213 is 11. Thus, 11 is the greatest number that satisfies the given conditions.

The number must be divisible by 3. Also, N = LCM(5, 7, 9) * k + 3 = 315k + 3. We check multiples of 315 plus 3 that are divisible by 3. 315k + 3 is always divisible by 3. We want the largest 4-digit number: 315k + 3 <= 9999 => 315k <= 9996 => k <= 31.7. For k=31, N = 315*31 + 3 = 9765 + 3 = 9768.

Statement A implies the number is 42k + 21, which is divisible by 7 (since 42 and 21 are multiples of 7). Statement B implies the number is a multiple of 98, which is also a multiple of 7. Thus, either statement alone is sufficient.