If the unit digits of N^33 is 7, which of the following could be the value of N?
a) 41
b) 43
c) 47
a) 41
b) 43
c) 47
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!
