${222}^{333}+{333}^{222}$ is divisible by which of the following numbers?

The correct option is 3 and 37 but not 2.

Let us analyze the divisibility of the expression $E={222}^{333}+{333}^{222}$ by 2, 3, and 37 step-by-step.

1. Divisibility by 2:
We examine the parity (even or odd nature) of each term in the sum:
The number 222 is even, so any positive integer power of it, ${222}^{333}$, is also even.
The number 333 is odd, so any positive integer power of it, ${333}^{222}$, is also odd.
The sum of an even number and an odd number is always odd:
$\text{Even}+\text{Odd}=\text{Odd}$
Since $E$ is an odd number, it is not divisible by 2.

2. Divisibility by 3:
Let us check the terms modulo 3:
For the first term, we sum the digits of 222: 2 + 2 + 2 = 6, which is divisible by 3. Thus, $222\equiv 0\left(\mathrm{mod}3\right)$. This means:
${222}^{333}\equiv 0^{333}\equiv 0\left(\mathrm{mod}3\right)$
For the second term, we sum the digits of 333: 3 + 3 + 3 = 9, which is divisible by 3. Thus, $333\equiv 0\left(\mathrm{mod}3\right)$. This means:
${333}^{222}\equiv 0^{222}\equiv 0\left(\mathrm{mod}3\right)$
Therefore, the sum modulo 3 is:
$E\equiv 0+0\equiv 0\left(\mathrm{mod}3\right)$
This confirms that the expression is divisible by 3.

3. Divisibility by 37:
Let us analyze the base numbers 222 and 333:
Notice that $222=6\times 37$ and $333=9\times 37$.
Since both bases are multiples of 37, we have:
$222\equiv 0\left(\mathrm{mod}37\right)$ and $333\equiv 0\left(\mathrm{mod}37\right)$
Raising these to their respective powers yields:
${222}^{333}\equiv 0^{333}\equiv 0\left(\mathrm{mod}37\right)$
${333}^{222}\equiv 0^{222}\equiv 0\left(\mathrm{mod}37\right)$
Thus, the sum modulo 37 is:
$E\equiv 0+0\equiv 0\left(\mathrm{mod}37\right)$
This confirms that the expression is divisible by 37.

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