Question

find the sum of all 3 didgit number divisible by 9​

Answer 1

Answer:

55350

Step-by-step explanation:

Answer 2

Answer:

49815

Step-by-step explanation:

The smallest 3 digit number divisible by 9 is 108.

The greatest 3 digit number divisible by 9 is 999.

Sum of all the terms of an arithmetic series =

$\frac{n}{2}(a + l)$

where a and l are the first and last terms respectively.

$l = a + (n - 1)d$

$999 = 108 + 9(n - 1)$

$999 - 108 = 9(n - 1)$

$801 \div 9 = n - 1$

$89 = n - 1$

$90 = n$

Now substitute the value of n in the first equation:

$\frac{90}{2} (108 + 999)$

$45(1107)$

$= 49815$