Number System

If the ratio is 1:2:3 and HCF is 23, the numbers are 23*1, 23*2, 23*3, which are 23, 46, 69.

Product of two numbers = LCM * HCF. 130 * x = 1820 * 26. x = (1820 * 26) / 130 = 1820 / 5 = 364.

Find GCD of (640-11), (710-7), (1526-9) = GCD(629, 703, 1517). 629 = 17 * 37. 703 = 19 * 37. 1517 = 41 * 37. The GCD is 37.

If a number N = 225k + 32, then N = 15(15k) + 15(2) + 2. Dividing by 15, the remainder is 2.

The number must divide the differences: 3155-1475 = 1680, 5255-3155 = 2100, 5255-1475 = 3780. GCD(1680, 2100, 3780) = 420.

The number must divide (2272 - R) and (875 - R) for some remainder R. This implies the number must divide the difference (2272 - 875) = 1397. Factoring 1397 gives 11 * 127. Thus, 127 is the divisor.

LCM = 36 * HCF. Product = LCM * HCF = 36 * HCF^2 = 3600. HCF^2 = 100 => HCF = 10.

The number n satisfies n = 11a + 7, n = 13b + 9, n = 17c + 3. This is a system of congruences. Solving this, n = 2570 is a solution. Checking: 2570/11 = 233 R 7, 2570/13 = 197 R 9, 2570/17 = 151 R 3.

The divisors are 8, 12, 15, 20. LCM(8, 12, 15, 20) = 120. The differences between divisors and remainders are constant: 8-5=3, 12-9=3, 15-12=3, 20-17=3. We need the largest 5-digit number N such that N = 120k - 3. 99999 / 120 = 833 with remainder 39. 120 * 833 = 99960. 99960 - 3 = 99957.

When any integer is divided by 4, the possible remainders are 0, 1, 2, and 3. Five consecutive integers will cycle through these remainders.

If the same remainder is obtained, the difference between the numbers (40718 - 40258 = 460) must be divisible by the divisor. The divisors could be any factor of 460. Since 115, 230, and 460 are all factors of 460, any of them could be the 3-digit number.

The product of two numbers is equal to the product of their HCF and LCM. Given HCF=37, LCM=1517, and one number=37, the other number is (37 * 1517) / 37 = 1517.

The product of two numbers equals the product of their HCF and LCM. 45 * x = 3 * 1035. x = (3 * 1035) / 45 = 1035 / 15 = 69.