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Question
Find the number of all three digit natural numbers which are divisible by 9.
Solution
First three-digit number that is divisible by 9 is 108.
Next number is 108 + 9 = 117.
And the last three-digit number that is divisible by 9 is 999.
Thus, the progression will be 108, 117, .... , 999.
All are three digit numbers which are divisible by 9, and thus forms an A.P. having first term a 108 and the common difference as 9.
We know that, nth term = an = a + (n − 1)d
According to the question,
999 = 108 + (n − 1)9
⇒ 108 + 9n − 9 = 999
⇒ 99 + 9n = 999
⇒ 9n = 999 − 99
⇒ 9n = 900
⇒ n = 100
Thus, the number of all three digit natural numbers which are divisible by 9 is 100.
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