Question

5. Find the number of all three digit natural numbers which are divisible by 9
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Answer 1

Answer:

100 is the answer

Here's the explanation:

The lowest number greater than a hundred and divisible by 9 is 108... the largest number divisible by 9 and less than a thousand is 999... 

we use this formula... 

T = A + (n-1)*d 

T 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) = 999 

common difference(d) = 9 

we substitute 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 

( 891 ) / 9 = [ ( n - 1 ) * 9 ] / 9 

99 = n - 1 

99 + 1 = n 

100 = n 

therefore... 

there are 100 3-digit numbers divisible by 9 

Hope this answer helpful to you...