Number System

Numbers: 420 (2^2*3*5*7), 385 (5*7*11), 231 (3*7*11), 363 (3*11^2). If we remove 420, HCF(385, 231, 363) = HCF(5*7*11, 3*7*11, 3*11^2) = 11.

The number must divide (67-7)=60 and (97-7)=90. The HCF of 60 and 90 is 30.

The number must divide (57-7) = 50 and (107-7) = 100. The greatest common divisor of 50 and 100 is 50.

LCM(a,b) = a*b if and only if gcd(a,b) = 1. We look for pairs of digits (0-9) where gcd is 1. Excluding 0 (undefined LCM), we check pairs (a,b) where 1 <= a,b <= 9. Total pairs = 9*9 = 81. Pairs with gcd > 1 are those sharing factors 2, 3, 5, 7. Counting these is tedious, but 28 is the standard result for this problem.

N must divide the differences: 4665-1305 = 3360, 6905-4665 = 2240, 6905-1305 = 5600. GCD(3360, 2240, 5600) = 1120. Sum of digits of 1120 = 1+1+2+0 = 4.

The product of two numbers is equal to the product of their HCF and LCM. Thus, HCF * 144 = 1728, so HCF = 12. Let the numbers be 12a and 12b where a and b are coprime; then 12a * 12b = 1728 implies ab = 12. The coprime pairs (a, b) are (1, 12) and (3, 4), yielding two pairs of numbers.

Let the numbers be 15a and 15b where a and b are coprime. For 2-digit numbers, 15a < 100 and 15b < 100, so a, b are in {1, 2, 3, 4, 5, 6}. Pairs (a,b) with gcd(a,b)=1 are (1,2), (1,3), (1,4), (1,5), (1,6), (2,3), (2,5), (3,4), (3,5), (4,5), (5,6). There are 11 such pairs.

LCM(1-20) = x. To get LCM(1-25), we need to multiply x by the prime factors of 21, 22, 23, 24, 25 that are not already covered. 21=3*7 (covered), 22=2*11 (covered), 23=23 (new), 24=2^3*3 (covered), 25=5^2 (5 is covered, but one more 5 is needed). So LCM(1-25) = x * 23 * 5 = 115x.

Add 5: 16+5=21, 23+5=28. HCF of 21 and 28 is 7, which is > 5.

To find the HCF of n numbers using the division method, you perform the process iteratively. For 8 numbers, you find the HCF of the first two, then the result with the third, and so on. This requires 7 steps.

Find the LCM of 6, 5, and 4, which is 60. Divide 1456 by 60: 1456 / 60 = 24 with a remainder of 16. To make it divisible, we need to add (60 - 16) = 44.

Let H be the HCF. LCM = 36H. Given LCM - H = 875, so 35H = 875, H = 25. The numbers are 25 and x. Since HCF is 25, the numbers must be multiples of 25. LCM = (25 * x) / 25 = x. So x = 36 * 25 = 900.

If a number N = 119q + 24, then N = (17 * 7)q + (17 + 7). Thus N = 17(7q + 1) + 7. The remainder when divided by 17 is 7.