Question

How many three-digit numbers are divisible by 9 ?

Give solution in detail that how u get the terms such as first term , last term, etc

Answer 1

Answer:

The first three digit number is 100. The first three digit number divisible by 9 is 108.

This can be obtained by adding 9 to the largest 2 digit number divisible by 9. Largest 2 digit number divisible by 9 is 99. Hence adding 9 we get, 108 as the three digit number which is the first term. Last three digit number is 999 and eventually it is divisible by 9. So last term is 999.

So, a = 108, d = 9, l = 999, n = ?

We know that,

l = a + ( n - 1 ) d

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

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

=> 891 = ( n - 1 ) 9

=> 891 / 9 = ( n - 1 )

=> 99 = n - 1

=> n = 99 + 1 = 100 terms

Hence there are 100 three digit terms between 108 and 999 which are divisible by 9.

Answer 2

$\huge \bold {\mathfrak {hey}}$

✔️ Answer

The first three digit number is 100 but the first three digit number which is divisible by 9 is 108 by 9 × 12 = 108 and last 3 digit number which is divisible by 6 is 999 by Multiply 9 × 111 = 999.

✔️✔️✔️

Now,

Therefore, a = 108, L = 999, n =?, d =9

So,

L = a + (n - 1)d
➡️ 999= 108 + (n-1)9
➡️ 891 = (n-1)9
➡️ 891/9 = (n-1)
➡️ 99 = n - 1
➡️ n = 99+1=100✔️✔️✔️

Therefore, 100 terms are of three digits which can be divisible by 9.


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