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