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We need to have an idea about the last digits of numbers when raised to successive powers.

Take 41. When 1 (the last digit) is raised to any power, the last digit will always remain 1.

Take3 now. 3(^1) gives 3 as last digit.
Take3 now. 3(^2) gives 9 as last digit.
Take3 now. 3(^3) gives 7 as last digit.
Take3 now. 3(^4) gives 1 as last digit.

Then this sequence repeats itself for every new set of 4 powers (3^5, 3^6, 3^7 and 3^8) and so on.

Coming to 7 now,

7 when raised to successive powers gives 7, 9, 3 and 1 and this sequence repeats itself.

Now lets come to the question.

We have units digit of N^33 is 7.

If 41 is the number, this is not possible as always the last digit will be 1.
If 43 is the number, then 7 does appear as the 3rd in every sequence of 4, i.e., 3^3, 3^7, 3^11...3^27, 3^31, 3^35. Notice that 3^33 is missed here, So, this is not correct.

Coming to 47, for the last digit 7, 7 is the last digit for 7^1, 7^5, 7^9, 7^13,7^17.....7^29 and 7^33.

So, this is correct.

Answer is C!

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