Find the least residue of 7^5 mod 50 without using a calculator.
so far i have 7=7 mod 50 7^5 = 7^5 mod 50 7^2 = 49 7^3=343 since 7^2=(-1)mod50 and 7^3=(-43)mod50 it follows that 7^6=7^2 * 7^3 = (-) = 43 feel like this is wrong though could somebody explain where i have made an error?

7^1 = 7 (mod 50)
7^2 = 49 = -1 (mod 50)
7^3 = 7^2*7 = (-1)*7 = -7 (mod 50)
7^4 = (7^2)*(7^2) = (-1)*(-1) = 1 (mod 50)
7^5 = (7^4)*7 = 1*7 = 7 (mod 50)

So 7^5 = 7 (mod 50)

In other words, if we divided 7^5 over 50, the remainder would be 7. We don't need to worry about the quotient. 

1. isolate the variable, in this case, d. subtract 35.5 on both sides of the equation.2. 52 subtracted by 35.5 is 16.53. therefore your answer would be d= 16.5this is the basics of pre algebra, isolating variable through adding, subtracting, multiplying, and dividing, if you have any other questions to ask just comment!