Question

digit number find the number of all three digits number which is divisible by 9​

Answer 1

Answer:

answer is 3 ............

Answer 2

Answer:

$\red{ \sf The \;number \;of \;all \;three-digit \;natural \;numbers \;which \;are \;divisible\; by \;9 \;is \;\it 100.}$

Step-by-step explanation:

Solution

Three digit numbers that are divisible by 9 are:-

108,117,126,135.....999.

This is an AP whose ,

First term (a) = 108

     and

Common difference(d) = a2-a1

Here,

a1 = 108 and a2 = 117

$\implies$ d = 117-108 = 9

We know that nth term a(n) of an AP is given by the equation,

an = a+ (n-1)d

Here,

a(n) or the last term = 999

a = 108

d = 9

∴ 999 = 108 + (n-1)9

$\implies$ 999 = 108+ 9n-9

$\implies$ 999 = 99+9n

$\implies$ 999-99 = 9n

$\implies$ 900  = 9n

n = 100

To Verify

a(n) = a+(n-1)d

a(n) = 108+ (100-1)9

a(n) = 108 +99 ×9

a(n) = 108+ 891

a(n) = 999

Hence the answer is correct.