Number System

Let numbers be 12a and 12b where gcd(a,b)=1. Product = 144ab = 6336. ab = 44. Pairs (a,b) such that gcd(a,b)=1 are (1,44) and (4,11). Total 2 pairs.

LCM = 1001, HCF = 7. Numbers are 7a and 7b where a, b are coprime. 7 * a * b = 1001, so a * b = 143. Pairs (a, b) such that gcd(a, b) = 1 are (1, 143) and (11, 13). Thus, 2 pairs are possible.

The product of two numbers must be a multiple of the square of their HCF if the numbers are coprime multiples of the HCF. More simply, the product must be divisible by HCF^2 = 48^2 = 2304. 4608 / 2304 = 2, which is possible.

Given HCF = 11, the numbers are 11x and 11y where x and y are coprime. Their product is 11x * 11y = 11 * 693, so xy = 63. Their sum is 11(x + y) = 176, so x + y = 16. The pairs of factors of 63 that sum to 16 are 7 and 9. Thus, the numbers are 11*7 = 77 and 11*9 = 99.

For any two numbers, HCF * LCM = Product of the numbers. Here, 50 * 20 = 1000.

The product of two numbers equals the product of their HCF and LCM. 864 * x = 96 * 1296. x = (96 * 1296) / 864 = 144.

The number must divide (1657 - 6) = 1651 and (2037 - 5) = 2032. Find GCD(1651, 2032). 1651 = 13 * 127. 2032 = 16 * 127. GCD is 127.

The number must divide 630-6=624 and 940-4=936. The greatest common divisor of 624 and 936 is 312.

The product of two numbers equals the product of their HCF and LCM. Therefore, the other number = (HCF * LCM) / given number = (97 * 64699) / 2231 = 6275803 / 2231 = 2813.

Find LCM(12, 15, 20, 54). 12=2^2*3, 15=3*5, 20=2^2*5, 54=2*3^3. LCM = 2^2 * 3^3 * 5 = 4 * 27 * 5 = 540. The number is LCM + remainder = 540 + 7 = 547.

The least natural number divisible by all digits from 1 to 9 is the least common multiple (LCM) of {1, 2, 3, 4, 5, 6, 7, 8, 9}. LCM(1..9) = 2^3 * 3^2 * 5 * 7 = 8 * 9 * 5 * 7 = 2520.

Product of two numbers = HCF * LCM. 64 * x = 16 * 192. x = (16 * 192) / 64 = 192 / 4 = 48.

We need the HCF of (258 - 2) and (323 - 3), which is the HCF of 256 and 320. 256 = 64 * 4 and 320 = 64 * 5. The HCF is 64.

LCM = 1530, HCF = 51. Numbers are 51a and 51b where a, b are coprime. 51a * 51b / 51 = 1530 => 51ab = 1530 => ab = 30. Pairs (a,b) such that gcd(a,b)=1: (1,30), (2,15), (3,10), (5,6). There are 4 pairs.

Numbers are 1630-6=1624, 525-3=522, 1280-4=1276. HCF(1624, 522, 1276). 522 = 2 * 3^2 * 29. 1624/29 = 56. 1276/29 = 44. HCF is 58.