Answer:
To prove that 2^55 + 1 is divisible by 33, we can use the fact that 2^5 = 32 ≡ -1 (mod 33). Therefore, 2^55 ≡ (-1)^11 = -1 (mod 33). Adding 1 to this, we get 2^55 + 1 ≡ 0 (mod 33), which means it is divisible by 33.
To prove that 313^3 - 313 × 299^2 is divisible by 7, we can use the fact that 313 ≡ 3 (mod 7) and 299 ≡ 5 (mod 7). Therefore, 313^3 - 313 × 299^2 ≡ 3^3 - 3 × 5^2 ≡ 27 - 3 × 25 ≡ 27 - 75 ≡ -48 ≡ 0 (mod 7), so it is divisible by 7.
To prove that 11^100 - 1 is divisible by 100, we can use the fact that 11 ≡ 1 (mod 100). Therefore, 11^100 ≡ 1^100 ≡ 1 (mod 100). Subtracting 1 from this, we get 11^100 - 1 ≡ 1 - 1 ≡ 0 (mod 100), so it is divisible by 100.
To prove that 10^6 - 5^7 is divisible by 59, we can use Fermat's Little Theorem, which states that if p is a prime number and a is not divisible by p, then a^(p-1) ≡ 1 (mod p). Here, 59 is a prime number. We have 10^6 ≡ 1 (mod 59) and 5^7 ≡ 5 (mod 59). Therefore, 10^6 - 5^7 ≡ 1 - 5 ≡ -4 ≡ 55 (mod 59), so it is divisible by 59.
To prove that 54^3 - 24^3 is divisible by 1080, we can factorize the expression as (54 - 24)(54^2 + 54 × 24 + 24^2). Both 54 - 24 and 54^2 + 54 × 24 + 24^2 are divisible by 1080, so the entire expression is also divisible by 1080.
mark me as brainliest thanks
