Maximum Possible Ratio With Digit Counting Function

Calculus Level 2

Let $d(n)$ denote the number of digits of $n$ . For example, $d(1929) = 4$ and $d(301) = 3$ .

What is the maximum value of the ratio $\dfrac{\big(d(n)\big)^6}{n}$ as $n$ ranges over the positive integers?

7.\overline{33}

7.29

6.4

7.76

1 solution

Laurent Shorts
Mar 6, 2016

With $d(n)$ constant, the maximum value is given for the smallest $n$ , which is $n=100...0=10^{d(n)-1}$ .

Let's find the maximum value for $\frac{d(n)^6}{10^{d(n)-1}}$ . Let $f(x)=\frac{x^6}{10^{x-1}}$ , we have then $f'(x)=\frac{6x^5·10^{x-1}-x^6·10^{x-1}·\ln(10)}{(10^{x-1})^2}=\frac{x^5·10^{x-1}(6-x·\ln(10))}{10^{2x-2}}$ .

$f'(x)=0 \leftrightarrow x=\frac{6}{\ln(10)}=2.606...$

Sign of $f'$ :

$x$	$\|$	1	$~$	2.606	$~$
$f'(x)$	$\|$	$~$	$+$	$~~~0$	$-$
$f(x)$	$\|$	$~$	$\nearrow$	$~$	$\searrow$

So max value is for $x=2$ or $x=3$ . $f(2)=\frac{2^6}{10}=6.4$ , $f(3)=\frac{3^6}{100}=7.29$ . Max value is 7.29.

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A Former Brilliant Member - 3 years, 7 months ago

Please explain in simple words... I am confused

Mystic Dragon EX - 8 months, 2 weeks ago

Maximum Possible Ratio With Digit Counting Function

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