$57242^{9 \times 7 \times 5 \times 3 \times 1} = (57242)^{945}$ The unit digit of the resultant number only depends on the unit digit of the given number 57242, i.e. 2.

Now, we know that: $2^1 = 2$ $2^2 = 4$ $2^3 = 8$ $2^4 = 16$ (unit digit 6) $2^5 = 32$ (unit digit 2) And so on.

So, 2 has a cyclicity of 4. Exponent of any number ending in 2 will produce a number that will end in 2, 4, 8, or 6.

Now, $945 = 944 + 1$

944 is divisible by 4. So, the last digit of $(57242)^{945}$ will be the same as that of $(57242)^1$, which is 2.

Hence, option (A) is correct.