Question

which is th smallest 3 digit number which leave remainder 2 division by 9​

Answer 1

Answer:

101

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Answer 2

First of all we will see that our number is in the from of :

= 9x + 2

So we substitute x with 11 first

we get,

9×11+2 = 99+2 = 101

so this is our first number : 101

Now if we keep on adding 9 to it then we will get all three digit numbers that will leave a remainder 2 when we divide by 9.

So with a hit and trial approach the last number of the series will be

= 9x +2 = 1001

9x= 999

x= 111

so the series starts at x = 11 and ends at x = 111.

so the numbers in between are 101.

so there are 101 three digit numbers that leaves a remainder 2 when divided by 9 .

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