SOLUTION: How many integers from 0 to 100, inclusive, have a remainder of 2 when divided by 7?

Let n be such a number and k be the quotient when we divide
n by 7

Let's see how it would be if we divided n by 7 and got a quotient of k:

We'd have this:

    k
 7) n

Then to finish the division we'd multiply k by 7 getting 7k
and we'd write it under the n:

    k
 7) n
   7k

Then we'd draw a line:

    k
 7) n
   7k

Then we'd subtract

    k
 7) n
   7k
   n-7k

And so our remainder would be n-7k

So we set the remainder = 2

  n-7k = 2
     n = 2+7k     <-- we add 7k to both sides

Since we only want the values of n from 0 to 100,
we write the inequality

   0 ≦ n ≦ 100

   0 ≦ 2+7k ≦ 100    <-- we substitute 2+7k for n
  -2  -2       -2       <-- we subtract 2 from all three parts.   
  ---------------
  -2 ≦   7k ≦ 98

-2/7 ≦    k ≦ 14   <-- we divide all three parts by 7

So the integers that are 0 through 14, inclusive,
are all the values of k and the only ones we can substitute in 
n = 2+7k and have a remainder of 2.

Since we must count 0, the number of values is 15.

Edwin