Answer:
(36^5-6^9)(38^9-38^8) is divisible by both 2, 3, and 37, and hence it is divisible by 30 and 37.
Explanation:
To prove that (36^5-6^9)(38^9-38^8) is divisible by 30, we need to show that it is divisible by both 2 and 3.
To prove that it is divisible by 2, we note that (36^5-6^9) is even, since it contains at least one factor of 2. We also note that (38^9-38^8) is even, since it can be factored as 38^8(38-1), and 38-1 is even. Therefore, their product is even and divisible by 2.
To prove that it is divisible by 3, we note that (36^5-6^9) is divisible by 3, since the sum of its digits is 3+6+5+6+9 = 29, which is divisible by 3. We also note that (38^9-38^8) is divisible by 3, since the sum of its digits is 3+8+9 = 20, which is divisible by 3. Therefore, their product is divisible by 3.
To prove that (36^5-6^9)(38^9-38^8) is divisible by 37, we can use Fermat's Little Theorem, which states that if p is a prime number and a is an integer not divisible by p, then a^(p-1) is congruent to 1 mod p. Since 37 is a prime number and 6 is not divisible by 37, we have:
6^36 ≡ 1 (mod 37)
Therefore, we can rewrite the expression as:
(36^5-6^9)(38^9-38^8) ≡ (36^5-6^9)(1-38) (mod 37)
Simplifying further, we get:
(36^5-6^9)(1-38) ≡ (1-6^9)(1-38) (mod 37)
= (1-531441)(1-38)
= (531440)(-37)
Since 531440 is divisible by 37, we have shown that (36^5-6^9)(38^9-38^8) is divisible by 37.
Therefore, we have proved that (36^5-6^9)(38^9-38^8) is divisible by both 2, 3, and 37, and hence it is divisible by 30 and 37.