Number System

Given 5m = 84k + 30, we can divide by 5 to get m = 16.8k + 6. Since m is an integer, k must be a multiple of 5, say k = 5n. Then m = 84n + 6. We want to find the remainder of 8m when divided by 84. Substituting m, 8m = 8(84n + 6) = 672n + 48. Since 672 is a multiple of 84 (84 * 8 = 672), 8m = 84(8n) + 48. The remainder is 48.

B is the largest root of w^2 - 6w + 8 = 0, which factors to (w-4)(w-2)=0, so B=4. A is HCF of two primes, which is 1. Substituting: p = (1+3)^2 + 4 + 5 = 25; q = (1+2)^2 + 4 + 25 = 38; r = (1+4)^2 + 4 + 38 = 67; s = (1+5)^2 + 4 + 67 = 107. Numbers not divisible by 5 are q, r, and s.

LCM(16, 18, 20, 25) = 3600. The number is 3600k + 4. Testing multiples: 3600*5 + 4 = 18004. 18004 / 7 = 2572 with remainder 0. So 18004 is the number.

By Euler's totient theorem or checking powers modulo 125, the 100th power of an integer modulo 125 can only be 0 (if multiple of 5) or 1 (if coprime to 5).

Find LCM(16, 18, 20, 25) = 3600. The number is of the form 3600k + 4. Testing multiples: 3600*5 + 4 = 18004. Check divisibility by 7: 18004 / 7 = 2572. Since it is divisible by 7, 18004 is the answer.

The expression is evidently 9^100 divided by 8. Since 9 leaves remainder 1 when divided by 8, every positive power of 9 also leaves remainder 1.

Let N = 114k + 14. We want N mod 19. 114 = 19 * 6, so 114k is divisible by 19. N mod 19 = 14 mod 19 = 14.

If a number N = 45k + 21, then N = 15(3k) + 15 + 6. Dividing by 15 leaves a remainder of 6.

HCF * LCM = Product of two numbers. 65 * 975 = 325 * x. x = (65 * 975) / 325 = 195.

The prime factorizations are 5^2 * 3^3 and 4^2 * 3^2. Since 4^2 = 2^4, the factors are 5^2 * 3^3 and 2^4 * 3^2. The LCM is the product of the highest powers of all prime factors present: 2^4 * 3^3 * 5^2 = 16 * 27 * 25 = 10800.

A number x leaves a remainder of 2 when divided by 7 if x = 7k + 2. For k=5, x = 35 + 2 = 37.

N = 5k + 2. If k is even (k = 2m), N = 5(2m) + 2 = 10m + 2, so the remainder is 2. If k is odd (k = 2m + 1), N = 5(2m + 1) + 2 = 10m + 5 + 2 = 10m + 7, so the remainder is 7. Since the remainder depends on whether k is even or odd, there is no fixed remainder.