Question

If the counting numbers are written in order, what is the value of the 2015th digit written?

Answer 1

1*9 + 2*(90) + 3*(x)

Answer 2

1*9 + 2*(90) + 3*(x) = 2015

Answer 3

up to 608 would be 2013 digits, so 2 more digits would be 6.. 0? i think?

Answer 4

if you count 0 as a counting number then its 6 if not then its 0

Answer 5

Your answer is correct but can you be more specific and explain your answer a little better. 0 is correct not 6.

Answer 6

Explain how you got 0.

Answer 7

ok there are 9 one digit numbers, 90 two digit numbers, so you start with 1 * 9 and add to that 2 * 90, so i get 189

Answer 8

189 digits total

Answer 9

up to 99

Answer 10

now going past 99 we have 3 digit numbers, if you included all 900 of them then it would exceed 2015

Answer 11

1 * 9 + 2 * 90 + 3 * 900 is larger than 2015, so i know its somewhere in the three digit numbers

Answer 12

so i assign a variable to the number of three digit numbers, i assign x to it giving me

1 * 9 + 2 * 90 + 3 *x

Answer 13

set that equal to 2015, giving me, 1*9 + 2*90 + 3x = 2015, then i solve for x and i get 608 and some decimals, so if i plug 608 in for x i end up with 2013 digits, so i just need two more digits, so next number is 609 2nd digit of that is 0

Answer 14

theres probably a better way of solving it but thats the best way i could think of

Answer 15

okay i understand thank you

Answer 16

yw

Answer 17

Numbers 1 - 9 account for first 9 digits
Numbers 10 - 99 account for digits 10 through 189
Numbers 100 - 199 account for digits 190 through 489
Numbers 200 - 299 account for digits 490 through 789
Numbers 300 - 399 account for digits 790 through 1089
Numbers 400 - 499 account for digits 1090 through 1389
Numbers 500 - 599 account for digits 1390 through 1689
Numbers 600 - 699 account for digits 1690 through 1989
7007017027037047057067077 0 8709
^ ^
| |
digit no. 1990 digit no. 2015