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# Find the number of all three digit natural numbers which are divisible by 9.

To do:

We have to find the number of three-digit natural numbers which are divisible by 9.

Solution:

Let $n$ be the number of three-digit natural numbers which are divisible by $9$.

Let $a$ be the first term and $d$ be the common difference.

Multiples of 9 are $9, 18, ....., 99, 108, ....., 999, 1008, ......$

The first three-digit number divisible by $9$ is $108$.

This implies,

$a = 108, d = 9$, last term $a_n = 999$

$a_n = a + (n – 1) d$

$999 = 108 + (n – 1) \times 9$

$999- 108 = 9n – 9$

$9n = 891 +9$

$9n = 900$

$n=\frac{900}{9}$

$n=100$

Therefore, 100 three-digit natural numbers are divisible by 9.

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