Question
How many three-digit natural numbers are divisible by 9?
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Answer
The three-digit natural numbers divisible by 9 are 108, 117, 126, ..., 999.
Clearly, three number are in AP.
Here, a = 108 and d = 117 - 108 = 9
Let this AP contains n terms. Then,
$a_n = 999$
$\Rightarrow 108 + (n - 1) \times 9 = 999 [a_n = a + (n - 1)d]$
$\Rightarrow 9n + 99 = 999$
$\Rightarrow 9n = 999 - 99 = 900$
$\Rightarrow n = 100$
Hence, there are 100 three-digit numbers divisible by 9.
Clearly, three number are in AP.
Here, a = 108 and d = 117 - 108 = 9
Let this AP contains n terms. Then,
$a_n = 999$
$\Rightarrow 108 + (n - 1) \times 9 = 999 [a_n = a + (n - 1)d]$
$\Rightarrow 9n + 99 = 999$
$\Rightarrow 9n = 999 - 99 = 900$
$\Rightarrow n = 100$
Hence, there are 100 three-digit numbers divisible by 9.
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