Math, asked by pankhadevijaya, 2 months ago

The three digit possible natural number using each of digit 9 0 and 8 ones are​

Answers

Answered by trinidadandrei32
1

Answer:

HOW MANY DIGIT IS VISIBLE BY 9?

Step-by-step explanation:

Three digit numbers start from 100 and end at 999.

Smallest three-digit natural number which is divisible by 9 is 108.

Now from 108, if you repeatedly add 9 with 108(until you reach 999), then the result will be a three-digit natural number divisible by 9.

That is first you get 108+ 9=117

Then 117+ 9 = 126

Then 126 + 9 = 135

……

until you reach 999.

Logic is that, as 108 is divisible by 9 then (108+9×n) will also be divisible by 9.

Where n is the number of times you add 9 with 108.

Now, question is …. maximum how many times we can add 9 with 108 this way?

At last, 108 + 9×n will be equal to 999(as 999 is the largest 3-digit number divisible by 9).

108+ 9×n = 999

=> 9×n = 891

=> n = 891/9 = 99.

So, maximum 99 times you can add 9 with 108.

Thus there are 99 three-digit natural number after 108 which are multiple of 9.

Therefore

Including 108 there are 99+1= 100 number of 3-digit numbers which are divisible by 9.

Answer is 100.

Answered by bajiraochalke1
1

Answer:

Answer is =

4

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