Question

Consider the sequence of two digit numbers which leave a remainder 1 on divisible by 3

Answer 1

Answer:

Two digit numbers starts with 10 and ends with 99.

10 leaves remainder of 1

11 - X

12 - X

13 leaves remainder of 1

14 - X

15 - X

16 leaves remainder of 1

Step-by-step explanation:

97 leaves remainder of 1

(all the numbers divisible by 3) +1 would leave the reminder 1.

1–99 we have 33 numbers divisible by 3

1–9 we have 3 numbers divisible by 3

We have to include the one digit number 9 and exclude the two digit number 99, as the next will fall in/apart from our consideration.

So, it is 33(divisible by 3 in 1–99) - 3 (to exclude the one digit) + 1(add 10 or include one digit 9) - 1 (exclude 100 or remove two digit 99) = 30