Step 1: Check divisibility by 2

• 222 is even, so $222^{333}$ is even.
• 333 is odd, so $333^{222}$ is odd.
• The sum of an even number ($222^{333}$) and an odd number ($333^{222}$) is odd.

Thus, $222^{333} + 333^{222}$ is not divisible by 2.

Step 2: Check divisibility by 3 Using the property of divisibility by 3, a number is divisible by 3 if the sum of its digits is divisible by 3.

• Sum of digits of 222 = 2 + 2 + 2 = 6, so 222 is divisible by 3. Therefore, $222^{333}$ is divisible by 3.
• Sum of digits of 333 = 3 + 3 + 3 = 9, so 333 is divisible by 3. Therefore, $333^{222}$ is divisible by 3.

Since both terms $222^{333}$ and $333^{222}$ are divisible by 3, their sum is also divisible by 3.

Thus, $222^{333} + 333^{222}$ is divisible by 3.

Step 3: Check divisibility by 37 To check divisibility by 37, use modular arithmetic:

• 222 mod 37 = 222 - 37 × 6 = 222 - 222 = 0.
  Hence, 222 ≡ 0 mod 37, and $222^{333} \equiv 0 \mod 37$.

• 333 mod 37 = 333 - 37 × 9 = 333 - 333 = 0.
  Hence, 333 ≡ 0 mod 37, and $333^{222} \equiv 0 \mod 37$.

Since both terms are divisible by 37, their sum is also divisible by 37.

Thus, $222^{333} + 333^{222}$ is divisible by 37.

Final Answer: The expression $222^{333} + 333^{222}$ is divisible by 3 and 37 but not 2.