how many 3 digit natural numbers are divisible by 9
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9
First three-digit number that is divisible by 9 = 108
Next number = 108 + 9 = 117
Therefore, 108, 117, …
All are three digit numbers which are divisible by 9 and thus, all these are terms of an A.P. having first term as 108 and common difference as 9.
The maximum possible three-digit number is 999. When we divide it by 9, the remainder will be 0. Clearly, 999 − 0 = 999 is the maximum possible three-digit number that is divisible by 9.
The series is as follows.
108, 117, … 999
Let 999 be the nth term of this A.P.
a = 108
d = 9
a n = 999
n = ?
a n = a + (n − 1) d
999 = 108 + (n − 1) 9
891 = (n − 1) 9
(n − 1) = 99
n = 100
Therefore, 100 three-digit numbers are divisible by 9.
Next number = 108 + 9 = 117
Therefore, 108, 117, …
All are three digit numbers which are divisible by 9 and thus, all these are terms of an A.P. having first term as 108 and common difference as 9.
The maximum possible three-digit number is 999. When we divide it by 9, the remainder will be 0. Clearly, 999 − 0 = 999 is the maximum possible three-digit number that is divisible by 9.
The series is as follows.
108, 117, … 999
Let 999 be the nth term of this A.P.
a = 108
d = 9
a n = 999
n = ?
a n = a + (n − 1) d
999 = 108 + (n − 1) 9
891 = (n − 1) 9
(n − 1) = 99
n = 100
Therefore, 100 three-digit numbers are divisible by 9.
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4
there are 100 digits are divisible by 9
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