Question

How many 3 digit numbers are such that when divided by 7, leave a remainder 3 in each case.

Answer 1

The first such number can be found manually, and such numbers form an AP with the common difference of 7 and the first term of 101 all until 997
$101 \: 108 \: 115.......997$
We know that in an AP ,
$an = a + d(n - 1) \\ 997 = 101 + 7(n - 1) \\ 896 = 7(n - 1) \\ 128 = n - 1 \\ n = 129$
Thus, there are 129 3-digit numbers that leave a remainder of 3 when divided by 7