Question

What is the least three digit number, which is a multiple of 6?
Find the sum of all three digit numbers which are multiples of 6.

Answer 1

Least three digits are 100.
If we divide 100 by 6, we get remainder as 4.
Greatest two digit number divisible by 6 = 100 - 4 = 96
Least 3 digit number divisible by 6 = 96 + 6 = 102
Next number = 102 + 6 = 108
Therefore, the series is 102, 108, 114, ..........
The maximum possible three digit number is 999.
When we divide 999 by 6, the remainder will be 3.
Clearly, 999 - 3 = 996 is the maximum possible three digit number multiple of 6.
The series is as follows
102, 108, 114, .........., 996
Here a = 102, d = 6
Let 996 be the nth term of this A.P.
$a_n=a+(n-1)d$
$\Rightarrow 996=102+(n-1)6$
$\Rightarrow (n-1)6=894$
$\Rightarrow n-1=149$
$\Rightarrow n=150$
So there are 150 terms in this A.P.
Sum $=\frac{n}{2}$ (first term + last term)
$=\frac{150}{2}(102+996)$
$=75\times 1098$
$=82350$