MHB Find Max & Min for $\frac{n}{f(n)}$ when $n\in N$ and $9<n<100$

[Albert1]

if $n\in N$ , $9<n<100$, and $f(n)=$ all digits sum of $n$
(1)find $max(\dfrac {n}{f(n)})$ and $min(\dfrac {n}{f(n)})$
now if:
if $n\in N$ , $999<n<10000$, and $f(n)=$ all digits sum of $n$
(2)find $max(\dfrac {n}{f(n)})$ and $min(\dfrac {n}{f(n)})$

 

[acegikmoqsuwy]

My immediate guesses:

(1) Max is 10 and min is 19/10

(2) Max is 1000 and min is 1999/28

 

[Albert1]

My immediate guesses:

(1) Max is 10 and min is 19/10

(2) Max is 1000 and min is 1999/28

(1) both correct
(2) Max correct ,min incorrect
please write down the way you got those answers

 

[Opalg]

(1) both correct
(2) Max correct ,min incorrect
please write down the way you got those answers

[sp]My guess for the min in (2) was $1199/20 = 59.95$. But I haven't thought about how to prove it.[/sp]

 

[Albert1]

[sp]My guess for the min in (2) was $1199/20 = 59.95$. But I haven't thought about how to prove it.[/sp]

the min (2) is $\dfrac {1099}{19}$

 

[Opalg]

the min (2) is $\dfrac {1099}{19}$

I should have thought of that! (Fubar)

 

[Albert1]

if $n\in N$ , $9<n<100$, and $f(n)=$ all digits sum of $n$
(1)find $max(\dfrac {n}{f(n)})$ and $min(\dfrac {n}{f(n)})$
now if:
if $n\in N$ , $999<n<10000$, and $f(n)=$ all digits sum of $n$
(2)find $max(\dfrac {n}{f(n)})$ and $min(\dfrac {n}{f(n)})$

sol of (1)

Spoiler

let : $n=\overline{xy}=10x+y, (x\in 1...9 , y\in 0...9)$
$\therefore \dfrac {n}{f(n)}=\dfrac {10x+y}{x+y}=10-\dfrac {9y}{x+y}$
and we have :$max(\dfrac {n}{f(n)})=10 \,\, \,\,with (y=0, x=1...9)$
$min(\dfrac {n}{f(n)})=10 -\dfrac {9\times 9}{1+9}=\dfrac {19}{10}\,\,(y=9,x=1)$
sol of (2) is similar ,please have a try

 

[Albert1]

sol of (2) of others

Spoiler

let : $n=\overline{xyzw}=1000x+100y+10z+w, (x\in 1...9 , y,z,w\in 0...9)$
$\therefore \dfrac {n}{f(n)}=\dfrac {1000x+100y+10z+w}{x+y+z+w}=1000-\dfrac {900y+990z+999w}{x+y+z+w}$
and we have :$max(\dfrac {n}{f(n)})=1000 \,\, \,\,with (y,z,w=0, x=1...9)$
to find $min(\dfrac {n}{f(n)})= ?$
we must find $max(\dfrac {900y+990z+999w}{x+y+z+w})$
$\dfrac {900y+990z+999w}{x+y+z+w}
=900+\dfrac {90z+99w-900x}{x+y+z+w},\,\
( x=1,y=0)$
$=900+\dfrac {90z+99w-900}{1+z+w}$
$=900+90+\dfrac {9w-990}{1+z+w},\,\,(z=9)$
$=900+90+9-\dfrac {1080}{10+w}, \,\,\, (w=9)$
$=\dfrac {17901}{19}$
$\therefore min(\dfrac {n}{f(n)})=1000-\dfrac {17901}{19}=\dfrac {1099}{19}$#