Set up the original remainder equality. p will always be a positive integer. Look for any restrictions, like it having to be odd/even, or anything else. Then plug a value for p that meets all restrictions to get a sample x, plug the x back into the new number and find the remainder.
1. Twice the sum of 3 integers x ,y and z when divided by 7 gives remainder 1
what is the remainder when x+y+z is divided by 7?
realize that we'll always be working with (x+y+z) and never with the individual variables, so you should simplify the problem by substitution (S for x+y+z). That way, there's a greater chance that you'll "see" the solution.
set up the original problem:
2S=7p+1, where p is some multiple of 7 divided by 7. We know 2S must be even, and therefore p must be odd (do you see why?).
So now just plug in any odd number for p and you'll get a valid S. For example, use p=3:
2S=21+1=22
S=11
11/7 leaves remainder of 4.
clipped from: www.urch.com