asdert wrote:

I like johnrb's method. Mine wasn;t as fancy.

Did 2 lists. First, the column on the right with multiples of 7. Then a column on the right, adding 3 to the number on the left:

7 10

14 17

21 24

28 31

35 38

42 45

49 52

56 59

63 66

70 73

Then went though it to find a mutliple of five+1. Whole thing didn't take a minute to do.

Personally, when it comes to remainders, simpler is better. Lots of algebra looks great and sophisticated, but often is confusing and time consuming. I think this method, though not "as fancy", yields the right answer nearly immediately.

Note that asdert started with multiples of 7. That's the right move. S/He could have started with 5, but then there are just more numbers to write out on the list. So start with the higher number, then make the other number conform.