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

Correct Option (2)

The given expression is $222^{333} + 333^{222}$222333+333222.

To determine divisibility by 2:

• The term $222^{333}$222333 involves an even base (222) raised to a positive integer power, which results in an even number.
• The term $333^{222}$333222 involves an odd base (333) raised to a positive integer power, which results in an odd number.
• The sum of an even number and an odd number is always an odd number.
• Therefore, $222^{333} + 333^{222}$222333+333222 is an odd number and is not divisible by 2.

To determine divisibility by 3 and 37:

• The expression can be rewritten by factoring out 111 from the bases: $(2 \times 111)^{333} + (3 \times 111)^{222}$(2×111)333+(3×111)222.
• This can be further factored as $111^{222} (2^{333} \times 111^{111} + 3^{222})$111222(2333×111111+3222).
• Since $111^{222}$111222 is a factor of the entire expression, the expression is divisible by 111.
• The prime factorization of 111 is $3 \times 37$3×37.
• Consequently, the expression is divisible by both 3 and 37.

Based on this analysis, the expression is divisible by 3 and 37, but not by 2.

Incorrect Options:

Options 1, 3, and 4 are incorrect because the expression is an odd number, as determined by the sum of an even and an odd number. Therefore, it is not divisible by 2.

• Option 1 incorrectly states divisibility by 2.
• Option 3 incorrectly states divisibility by 2 and non-divisibility by 3.
• Option 4 incorrectly states divisibility by 2.