Help ill give u 25 points find the last digit of these numbers (and please explain why)

1)2^30
2)7^28
3)(5^2005)-5
4)(11^2005)-4^2005
5)(8^15)-3^7
6)3^75

Step-by-step explanation:

1) 2^1 = 2, 2^2=4, 2^3=8 2^4=6 2^5=2 .. we can find out a pattern: 2,4,8,6,2,4..

so 2^30 (mod 10) = 2^2 (mod 10) = 4

2) similar to Q1, 7^1 = 7, 7^2 = 9, 7^3 = 3, 7^4,=1, 7^5 = 7, the pattern is: 7,9,3,1,7.. so 7^28 = 7^4 = 1

3) since any power of 5 is equal to 5, (5^2005)-5 (mod 10) = 5-5 = 0

4) since any power of 1 is equal to 1, and there is a pattern for the power of 4 that is 4,6,4,6.. so (11^2005)-4^2005 = 1 - 4 (mod 10) = 7

5) 8's pattern is 8,4,2,6 and 3's pattern is 3,9,7,1, so (8^15)-3^7 = 2 - 7 = 5

6) 3's pattern is 3,9,7,1, so (3^75) (mod 10) = 7

step-by-step explanation:

a rhombus is a parallelogram enjoying his symmetry.

3 is greater then both of those numbers