Consider a set T={x, x+1, x+2, x+3, x+4, x+5}, where x can take any in

If x is one greater than some multiple of 7, then when we divide x by 7, the remainder will be 1, when we divide x+1 by 7, the remainder will be 2, and so on. Dividing all six numbers in our set by 7 will give us, in order, remainders of 1, 2, 3, 4, 5 and 6. So we won't have any multiples of 7 in our set in that case, but in any other case we will.

So the question is just asking, "how many positive integers less than 1000 give a remainder of 1 when you divide them by 7?" We can list the numbers we're talking about:

1, 8, 15, 22, .... , 981, 988, 995

and now to count how many things are in this list, since it is equally spaced, we can use the formula n = (range/space) + 1, so n = (995-1)/7 + 1 = 994/7 + 1 = 142 + 1 = 143. Or if we want to avoid formulas, we can just add 6 to everything in the list:

7, 14, 21, ...., 994, 1001

and divide everything by 7

1, 2, 3, ...., 142, 143

and we have a list just as long as our first list, which clearly has 143 things in it.