Question

If the numbers from 1 to 1000 which are exactly divisible by 3 are written in an ascending order, which would be the 100th digit from the left?​

Answer 1

Answer:

Step-by-step explanation:

3,6,9=3 dits(digits)          3

12-99=30x2dits                 60

still have 37dits

37÷3=12R1

102,105...135=36dits           36

.......135138

           ↑        =the 100dit    1

                                       _______

                                              100

Answer 2

Answer:

1

Step-by-step explanation:

1-digit numbers= 3

• Total number of digits=3

2-digit numbers= 12=3*4 till 99=3*33 total 30 numbers

• Total number of digits= 30*2=60

3-digit numbers= 102=3*34 till 999=3*333 total 300 numbers

we are looking for 100th digit from the left and we know that 63 of them formed by 1- and 2- digit numbers. Total digits we need to add:

• 100-63=37

• 12 of 3-numbers give us 36 and 13th number's first digit will be the 100th from the left

• 102+3*12=138 is the 13th 3-digit number and it'f first digit is 1 which is the 100th digit altogether

Answer is 100th digit from left is 1 (the first digit of 138)