7^1 = 7
7^2 = 49
7^3 = 343
7^4 = 2401
7^5 = 16807
7^6 = 117649
7^7 = 823543
7^8 = 5764801
7^9 = 40353607
7^10 = 282475249
7^11 = 1977326743
the last digit is repeating in a pattern of:
7,9,3,1,
7,9,3,1,
7,9,3,1,
to find out what the last digit is, you need to make a formula.
the formula would be taking the larger exponent and breaking it down into the smaller exponent
example:
suppose your number is 7^6.
6/4 = 2 and 7^2 = 49 so the last digit is 9.
look at 7^6 above and you can see that the last digit is 9.
suppose your number is 7^11.
11/4 = 2 with a remainder of 3.
7^3 = 343 so the last digit is 3.
look up 7^11 above and you can see that the last digit is 3.
you can use this formula for larger numbers even though you can't check them because the calculator doesn't carry enough digits.
I used excel and the highest I could go was 7^17 as shown below:
base exponent number last digit
7 1 7 7
7 2 49 9
7 3 343 3
7 4 2401 1
7 5 16807 7
7 6 117649 9
7 7 823543 3
7 8 5764801 1
7 9 40353607 7
7 10 282475249 9
7 11 1977326743 3
7 12 13841287201 1
7 13 96889010407 7
7 14 678223072849 9
7 15 4747561509943 3
7 16 33232930569601 1
7 17 232630513987207 7
7 18 9
7 19 3
7 20 1
7 21 etc.....
for a final check, try 7^16.
16/4 = 4 with a remainder of 0.
7^0 = 1 so the remainder should be 1.
it is, as shown in the table from excel above.
note that 16 / 4 = 4 with a remainder of 0. it also equals 3 with a remainder of 4 which confirms that the last digit is 1.
In your problem, the number is 7^2005
divide 2005 by 4 to get 501 with a remainder of 1.
7^1 = 7, so the last digit of 7^2005 = 1