Minimum & maximum

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Mehedi Hasan Nowshad
Posts:12
Joined:Sat Jun 13, 2015 1:46 pm
Location:Halishahar, Chittagong
Minimum & maximum

Unread post by Mehedi Hasan Nowshad » Sat Oct 22, 2016 8:58 pm

Let $p$,$q$,$r$ be positive integers such that $pq+qr+r=2015$ . Find the minimum & maximum value of $p+q+r$ .
"Failure is simply the opportunity to begin again, this time more intelligently."
- Henry Ford

Mehedi Hasan Nowshad
Posts:12
Joined:Sat Jun 13, 2015 1:46 pm
Location:Halishahar, Chittagong

Re: Minimum & maximum

Unread post by Mehedi Hasan Nowshad » Sat Oct 22, 2016 11:20 pm

My solution
The given equation is equivalent to,

$p+r= \dfrac{2015-r}{q} $
or, $p+q+r= \dfrac{2015-r}{q} + q$

Now let $a= \dfrac{2015-r}{q}$ and $b=q$. Then we have,

\[ p+q+r=a+b \]

with $q=b$, $r=2015-ab$ and $p=a+ab-2015$.

So now we've to actually find the minimum and maximum values of $a+b$.

As, $p>0$, $q>0$, $r>0$, we have,

$a+ab-2015>0$, $b>0$ and $2015-ab>0$

Or, $a+ab>2015$ and $2015>ab$

So, $a+ab>2015>ab$

So, we see that if 'a' is not a divisor of 2015, then 2015 is equal to exactly one of the terms (ab+1,ab+2,....ab+a-1). In other words, we can say that if 'a' is fixed and $2015=ab+k$, then,

$b=\dfrac{2015-k}{a}=\left \lfloor \dfrac{2015}{a} \right \rfloor$ [$b$ is an integer]

So, $b=\left \lfloor \dfrac{2015}{a} \right \rfloor$ if $a$ is not a divisor of 2015 and otherwise $b$ has no possible values.
So now our problem has been reduced to find the minimum and maximum values of $a+\left \lfloor \dfrac{2015}{a} \right \rfloor$ where $a$ is not a divisor of 2015.
Again as $ab=a.\left \lfloor \dfrac{2015}{a} \right \rfloor \approx 2015$, the sum $a+b$ gets bigger as the difference between $a$ and $b$ gets higher.

So, maximum value occurs when $a$ or $b$ is equal to 2 (1 is not possible as 1 divides 2015). Then we have,

$max(p+q+r)=max(a+b)=2+\left \lfloor \dfrac{2015}{2} \right \rfloor =1009$.

And minimum value occurs when a or b is approximate to $\sqrt{2015} \approx 44 or 45$.

So, $min(p+q+r)=min(a+b)=44+\left \lfloor 2015/44 \right \rfloor = 89$
"Failure is simply the opportunity to begin again, this time more intelligently."
- Henry Ford

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