Answer:
The given expression is divisible by both 30 and 37
Explanation:
First, let's consider the expression (36^5-6^9). We can factor out 6^5 from both terms to get:
(36^5-6^9) = 6^5(6^10-36^3)
Next, let's consider the expression (38^9-38^8). We can factor out 38^8 from both terms to get:
(38^9-38^8) = 38^8(38-1)
Now, we can substitute these factorizations back into the original expression:
(36^5-6^9)(38^9-38^8) = 6^5(6^10-36^3)38^8(38-1)
To show that this expression is divisible by 30, we need to show that it is divisible by both 2 and 3. We can see that 6^5 is divisible by both 2 and 3, so the entire expression is divisible by 2 and 3, and hence divisible by 30.
To show that this expression is divisible by 37, we can use Fermat's Little Theorem, which states that if p is a prime number and a is any integer not divisible by p, then a^(p-1) is congruent to 1 mod p. In this case, p=37 and a=6, so we can write:
6^36 ≡ 1 (mod 37)
Multiplying both sides by 6^10 gives:
6^46 ≡ 6^10 (mod 37)
We can use this congruence to simplify the expression we are interested in:
(36^5-6^9)(38^9-38^8) ≡ (6^10-6^9)(1-38^-1) (mod 37)
Simplifying this expression further gives:
(6^10-6^9)(1-38^-1) ≡ 0 (mod 37)
Therefore, the expression (36^5-6^9)(38^9-38^8) is divisible by both 30 and 37.