222333 + 333222 is divisible by which of the following numbers?

A. 2 and 3 but not 37
B. 3 and 37 but not 2
C. 2 and 37 but not 3
D. 2, 3 and 37

Answer: B. 3 and 37 but not 2

Explanation:

Step 1: Check divisibility by 2

  • 222 is even, so 222333 is even.
  • 333 is odd, so 333222 is odd.
  • The sum of an even number (222333) and an odd number (333222) is odd.

  • Thus, 222333 + 333222 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, 222333 is divisible by 3.
  • Sum of digits of 333 = 3 + 3 + 3 = 9, so 333 is divisible by 3. Therefore, 333222 is divisible by 3.

  • Since both terms 222333 and 333222 are divisible by 3, their sum is also divisible by 3.

  • Thus, 222333 + 333222 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 222333 ≡ 0 mod 37.

  • 333 mod 37 = 333 - 37 × 9 = 333 - 333 = 0.
  • Hence, 333 ≡ 0 mod 37, and 333222 ≡ 0 mod 37.

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

  • Thus, 222333 + 333222 is divisible by 37.

Final Answer:

The expression 222333 + 333222 is divisible by 3 and 37 but not 2.