3 to a power divisible by 4 will have a units digit of 1.
The powers of 3 are 3, 9, 27, 81 ... obviously, the next one will have a units digit of 1x3 or 3, the next one will have a units digit of 3x3 or 9, the next one will have a units digit of 7 (because 9x3 is 27), the next one will have a units digit of 1 (because 7x3 is 21), and then the cycle starts over with a units digit of 3 again.
It is 1.
Here's an example. In the number 382, the number 2 is the "unit's digit" (in the "unit's place"), 8 is the "ten's digit" (in the "ten's place"), and 3 is the "hundred's digit."
it is 3
It is 63.
There are no such numbers.
7
The unit digit of 3127173 is the unit digit of 7173. The other digits of 3127 are multiples of 10 and so they cannot contribute to the unit digit. Now the unit digits of the powers of 7 are Power -- Unit digit 0 -- 1 1 -- 7 2 -- 9 3 -- 3 4 -- 1 and you are back into the loop (of 1-7-9-3). So, you only need consider 7 to the power 173 modulo 4. That is, the remainder when 173 is divided by 4. 173 = 1 mod 4 So the unit digit of 3127173 is the same as the unit digit of 7173 which is the unit digit of 71 which is 7.
Any digit in the tens or higher place has no influence on the answer. So it is the unit digit of 4*9*3*6 = unit digit of 6*3*6 = unit digit of 8*6 = 8
It is the 3.
The last digit is 3. If you multiply 3 repeatedly by itself, you get the last digits:3, 9, 7, 1, 3, 9, 7, 1... In other words, every fourth power of 3 will end with the digit 1. Since any factorial starting at 4! has 4 as a factor, it follows that the last digit is 1.
Here's an example. In the number 382, the number 2 is the "unit's digit" (in the "unit's place"), 8 is the "ten's digit" (in the "ten's place"), and 3 is the "hundred's digit."
3 13 23
3
3
it is 3
It is 63.
60
There are no such numbers.