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

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

Mathematics
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1*9 + 2*(90) + 3*(x)
1*9 + 2*(90) + 3*(x) = 2015
up to 608 would be 2013 digits, so 2 more digits would be 6.. 0? i think?

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if you count 0 as a counting number then its 6 if not then its 0
Your answer is correct but can you be more specific and explain your answer a little better. 0 is correct not 6.
Explain how you got 0.
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
189 digits total
up to 99
now going past 99 we have 3 digit numbers, if you included all 900 of them then it would exceed 2015
1 * 9 + 2 * 90 + 3 * 900 is larger than 2015, so i know its somewhere in the three digit numbers
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
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
theres probably a better way of solving it but thats the best way i could think of
okay i understand thank you
yw
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

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