Number System

x = 15k + 1. We want the remainder when x is divided by 5. x = 5(3k) + 1. The remainder is 1.

Given N = 23k + 19. We need the remainder of (2N + 1) / 23. Substituting N, we get 2(23k + 19) + 1 = 46k + 38 + 1 = 46k + 39. Dividing 39 by 23, the remainder is 39 - 23 = 16.

Let the numbers be 15x and 15y, where x and y are coprime. 15x + 15y = 150 implies x + y = 10. Possible pairs (x, y) are (1, 9), (3, 7). If (1, 9), numbers are 15 and 135. If (3, 7), numbers are 45 and 105. Since both must be greater than 15, 45 and 105 is the correct pair.

The LCM of 8 and 9 is 72. The largest 5-digit number is 99999. Dividing 99999 by 72 gives 1388 with a remainder of 63. The largest multiple of 72 less than 99999 is 99999 - 63 = 99936. To get a remainder of 15, we add 15 to this multiple: 99936 + 15 = 99951.

Let the number be N = 238k + 79. We want N mod 17. Since 238 = 17 * 14, 238k is divisible by 17. So N mod 17 = 79 mod 17. 79 = 17 * 4 + 11. Remainder is 11.

Let the numbers be x, x+1, x+2, x+3, x+4. LCM(x, x+1, x+2, x+3, x+4) = 60. Testing consecutive integers, 2, 3, 4, 5, 6 gives LCM 60. Product of first two is 2 * 3 = 6, which is the 5th number. Sum = 2+3+4+5+6 = 20.

Numbers are of the form 3k + 2. Smallest 3-digit number: 101 (3*33+2). Largest < 500: 497 (3*165+2). This is an AP: 101, 104, ..., 497. Number of terms n = (497 - 101)/3 + 1 = 396/3 + 1 = 132 + 1 = 133. Sum = n/2 * (first + last) = 133/2 * (101 + 497) = 133/2 * 598 = 133 * 299 = 39767.

If n divides 2616, 6951, and 9552 with the same remainder, n must divide the differences: 6951 - 2616 = 4335, 9552 - 6951 = 2601, and 9552 - 2616 = 6936. The HCF of 4335, 2601, and 6936 is 17^2 = 289.

The number N is a repeating sequence of '323' repeated. 323 divided by 19 is exactly 17 (19 * 17 = 323). Since N consists of 333 blocks of '323' plus two extra digits '32', the remainder is 32 mod 19, which is 13.

The numbers must divide the differences: 5880-3920=1960, 7350-5880=1470, 7350-3920=3430. The GCD of 1960, 1470, 3430 is 70. The number of divisors of 70 (2*5*7) is (1+1)(1+1)(1+1) = 8. However, the question asks for natural numbers leaving the same remainder, which involves divisors of the GCD.

We want to find n such that 1111 - 99n < 99. Solving 1111 - 99 < 99n gives 1012 < 99n, so n > 1012/99, which is approximately 10.22. The smallest integer n is 11.

If a number is divisible by 6, it is a multiple of 6 (6, 12, 18, 24, 30, 36...). Dividing these by 9: 6/9 rem 6, 12/9 rem 3, 18/9 rem 0, 24/9 rem 6, 30/9 rem 3. The possible remainders are 0, 3, 6.

The Highest Common Factor (HCF) of any two numbers must always be a factor of their Least Common Multiple (LCM). Among the given options, only 16 is a factor of 144 (144 / 16 = 9), so 16 can be their HCF.