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

Correct Option (2)

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

To determine divisibility by 2:

• The term ${222}^{333}$ involves an even base (222) raised to a positive integer power, which results in an even number.
• The term ${333}^{222}$ 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, ${222}^{333}+{333}^{222}$ 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: $(2\times 111{)}^{333}+(3\times 111{)}^{222}$.
• This can be further factored as ${111}^{222}(2^{333}\times {111}^{111}+3^{222})$.
• Since ${111}^{222}$ is a factor of the entire expression, the expression is divisible by 111.
• The prime factorization of 111 is $3\times 37$.
• 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.