taghdh9

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?

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?

assumintdz

Beginner2022-06-29Added 22 answers

In general, if a positive integer $n$ and $d$ digits in its decimal expansion then

$d-1\le \mathrm{log}(n)\le d$

For example $\mathrm{log}(1529)=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 $9\times 1+90\times 2+900\times 3+1014\times 4=6945$ digits.

So we must have

$6944\le \mathrm{log}(12345678\dots 2013)\le 6945$

and so $\mathrm{log}(6944)\le \mathrm{log}(\mathrm{log}(12345678\dots 2013))\le \mathrm{log}(6945)$

Can you take it from here?

$d-1\le \mathrm{log}(n)\le d$

For example $\mathrm{log}(1529)=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 $9\times 1+90\times 2+900\times 3+1014\times 4=6945$ digits.

So we must have

$6944\le \mathrm{log}(12345678\dots 2013)\le 6945$

and so $\mathrm{log}(6944)\le \mathrm{log}(\mathrm{log}(12345678\dots 2013))\le \mathrm{log}(6945)$

Can you take it from here?

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