taghdh9

## Answered question

2022-06-28

How would you fin the integer closest to
log(log(1234567891011121314...2013))
where the number is the concatenation of numbers 1 through 2013 inclusive. log() in this case is log base 10.
Also, how would you find the remainder when it is divided by 75?

### Answer & Explanation

assumintdz

Beginner2022-06-29Added 22 answers

In general, if a positive integer $n$ and $d$ digits in its decimal expansion then
$d-1\le \mathrm{log}\left(n\right)\le d$
For example $\mathrm{log}\left(1529\right)=3.1844\dots$ and $1529$ has $4$ digits.
How many digits are in $12345678\dots 2013$ ? Well we have
$9$ one-digit numbers
$99-9=90$ two-digit numbers
$999-99=900$ three-digit numbers
$2013-999=1014$ four-digit numbers
So your number has $9×1+90×2+900×3+1014×4=6945$ digits.
So we must have
$6944\le \mathrm{log}\left(12345678\dots 2013\right)\le 6945$
and so $\mathrm{log}\left(6944\right)\le \mathrm{log}\left(\mathrm{log}\left(12345678\dots 2013\right)\right)\le \mathrm{log}\left(6945\right)$
Can you take it from here?

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