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## Question

Starting with 1 which number contains the 1000th consecutive digit?

i.e

1 contains the first digit

2 contains the second digit

...

10 contains the 10th and 11th digits

11 contains the 12th and 13th digits

...

what is the highest consecutive digit that 1000 contains?

given a number n, what is the highest consecutive digit that it contains?

## 4 answers to this question

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Which number contains the 1000th consecutive digit?

370. If my calculation is correct, the 1000th digit '3' in 370.

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what is the highest consecutive digit that 1000 contains?

000-999 ==> 3*1000 digits = 3000 digits.

But then we need to subtract the leading (one) zeros in 000-099 ==> 100 digits. 00-999 ==> 3000-100=2900 digits.

We need to subtract the leading zeros in 00-09 ==> 10 digits. Then 0-999 ==> 2900-10=2890 digits.

OP started from 1. Subtracting the first zero: 1-999 ==> 2890-1=2889 digits. So the last digit in 999 is the 2889th digit.

1000 contains 2890th, 2891st, 2992nd and 2993rd digits.

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There are 9 single digit numbers, 90 2-digit numbers, 900 3-digit numbers, etc.

370

9*1 + 90*2 + x*3 = 1000

3x + 189 = 1000

3x = 811

x = 270.3333

Meaning that the 270th 3-digit number (369), will only have 999 total digits so far. The number 370 contains the 1000th digit (the number 3 is the 1000th)

2993

9*1+90*2+900*3 = 2889

1000 has 4 digits, so 2889+4 = 2993

Sigma notation for k going from 1 to d-1 of 9*d*10^(k-1) [end sigma notation] + d*(number - 10^(d-1)) where d is the number of digits the number has.

Let d = number of digits the number has

For x = 1 to d-1, calculate the sum of d*9*10^(x-1) for all values of x [calculates the total number of digits for smaller digit numbers]

Then, add that to d*(number-10^(d-1)) [calculates the total of number of digits with that number and all other numbers smaller with the same digits]

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omg that is cazy

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