Question

Find the number of all three digit natural numbers which are divisible by 9.

Answer 1

Solution

100 is the answer

The lowest 3 digit number divisible by 9 is 108.
The largest number divisible by 9 and less than a thousand because thousand becomes a four digit number is 999...
We use this formula...

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

Here $T_n$ is the nth term in an arithmetic sequence
a is the first term in an arithmetic sequence
n is the number of terms
d is the common difference

you are asking for "the number of 3 digit numbers divisible by 9" or numbers that are greater than 100 but less than 1000 that are divisible by 9

so...

we are looking for the number of terms given

first term(a) = 108
last term$T_n$ = 999
common difference(d) = 9

we subtitute the given values
999 = 108 + ( n - 1 ) * 9

we then solve...
999 = 108 + ( n - 1 ) * 9
999 - 108 = ( n - 1 ) * 9
891 = ( n - 1 ) * 9
$\frac{891}{9}$ = [ ( n - 1 ) * 9 ] / 9
99 = n - 1
99 + 1 = n
100 = n

Therefore., there are 100, 3-digit numbers divisible by 9