BDMO National Junior 2016/6

[Math Mad Muggle]

p= 3^w,q=3^x, r = 3^y, s = 3^z...(w,x,y,z are positive integer) Find the minimum value of (w+x+y+z) such that( p^2 + q^3 + r^5 = s^7)

[Kazi_Zareer]

$w+x+y+z=106$

[Math Mad Muggle]

But how?....Please explain it.

[dshasan]

$3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$
WLOG, Let's assume, $3^{2w} < 3^{3x} < 3^{5y}$
Then, $3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$
$\Rightarrow 3^{2w} ( 1 + 3^{3x-2w} + 3^{5y-2w}) = 3^{7z}$
$\Rightarrow 1 + 3^{3x-2w} + 3^{5y-2w} = 3^{7z-2w}$
Which gives that R.H.S is divisible by $3$, but L.H.S is not unless $3^{3x-2w} = 3^{5y-3x} = 1.$ But that means $3^{2w} = 3^{3x} = 3^{5y}$, a contradiction.
Now, WLOG, Let's assume $3^{2w} = 3^{3x} < 3^{5y}$. Which gives us ,
$\Rightarrow 1 + 1 + 3^{5y-2w} = 3^{7z-2w}$.
But here also, R.H.S is divisible by $3$, but the L.H.S is not unless $3^{5y-2w} = 1$, another contradiction.
So, $3^{2w} = 3^{3x} = 3^{5y}$ , which gives us $7z - 2w = 7z - 3x = 7z - 5y = 1$
Solving the three equations, we get the least form of $w, x, y, z$ , which are $z = 13, w = 45, x = 30, y= 18$
So, $w + x + y + z = 13 + 45 + 30 + 18 = 106$

[Akhiar]

What do yo mean by WLOG
?

[Thamim Zahin]

It is no my solution. The credit goes to @thanicsamin

$3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$

Now, write this down in trinary form. It would be

$ 1\overbrace{000 \cdots 000}^{2^{2w}} + 1\overbrace{000 \cdots 000}^{2^{3x}} + 1\overbrace{000 \cdots 000}^{2^{5y}} = 1\overbrace{000 \cdots 000}^{2^{7x}} $

Now, If all the variables in LHS are not equal then the RHS would have something like $1000 \cdots 1000 \cdots 1000$ or something like that. But this is'nt true .So that means. all the variables of LHS is equal.

So, we get $3^{2w} = 3^{3x} = 3^{5y} $

By this we can easily get that $3^{2w} = 3^{3x} = 3^{5y} = 3^{7z-1}$

Or, $2w=3x=5y=7z-1$

So. $w+x+y+z=45+30+18+13=106$

[ahmedittihad]

$3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$
WLOG, Let's assume, $3^{2w} < 3^{3x} < 3^{5y}$
Then, $3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$
$\Rightarrow 3^{2w} ( 1 + 3^{3x-2w} + 3^{5y-2w}) = 3^{7z}$
$\Rightarrow 1 + 3^{3x-2w} + 3^{5y-2w} = 3^{7z-2w}$
Which gives that R.H.S is divisible by $3$, but L.H.S is not unless $3^{3x-2w} = 3^{5y-3x} = 1.$ But that means $3^{2w} = 3^{3x} = 3^{5y}$, a contradiction.
Now, WLOG, Let's assume $3^{2w} = 3^{3x} < 3^{5y}$. Which gives us ,
$\Rightarrow 1 + 1 + 3^{5y-2w} = 3^{7z-2w}$.
But here also, R.H.S is divisible by $3$, but the L.H.S is not unless $3^{5y-2w} = 1$, another contradiction.
So, $3^{2w} = 3^{3x} = 3^{5y}$ , which gives us $7z - 2w = 7z - 3x = 7z - 5y = 1$
Solving the three equations, we get the least form of $w, x, y, z$ , which are $z = 13, w = 45, x = 30, y= 18$
So, $w + x + y + z = 13 + 45 + 30 + 18 = 106$

Your solution is correct, but you can't say wlog here. It doesn't matter much. But there may be points taken from that.

[dshasan]

$3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$

Let's assume, $3^{2w} < 3^{3x} < 3^{5y}$

Then, $3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$

$\Rightarrow 3^{2w} ( 1 + 3^{3x-2w} + 3^{5y-2w}) = 3^{7z}$

$\Rightarrow 1 + 3^{3x-2w} + 3^{5y-2w} = 3^{7z-2w}$

Which gives that R.H.S is divisible by $3$, but L.H.S is not unless $3^{3x-2w} = 3^{5y-3x} = 1.$ But that means $3^{2w} = 3^{3x} = 3^{5y}$, a contradiction.

Same way, we can prove it for $3^{3x} < 3^{2w} < 3^{7z}$, $3^{7z} < 3^{2w} < 3^{3x}$ and such other cases.

Now, Let's assume $3^{2w} = 3^{3x} < 3^{5y}$. Which gives us ,

$\Rightarrow 1 + 1 + 3^{5y-2w} = 3^{7z-2w}$.

But here also, R.H.S is divisible by $3$, but the L.H.S is not unless $3^{5y-2w} = 1$, another contradiction.

Same way, we can prove it for $3^{3x} = 3^{7z}$ and $3^{2w} = 3^{7z}$

So, $3^{2w} = 3^{3x} = 3^{5y}$ , which gives us $7z - 2w = 7z - 3x = 7z - 5y = 1$

Solving the three equations, we get the least form of $w, x, y, z$ , which are $z = 13, w = 45, x = 30, y= 18$

So, $w + x + y + z = 13 + 45 + 30 + 18 = 106$

I hope it's correct now.

[Thanic Nur Samin]

Let's assume, $3^{2w} < 3^{3x} < 3^{5y}$

All you did was drop the word WLOG. You can't just assume that.

The correct way to write that is as follows: let $\{3^{2w},3^{3x},3^{5y}\}=\{3^{\alpha},3^{\beta},3^{\gamma}\}$.

where $\alpha < \beta < \gamma$.

[samiul_samin]

What do yo mean by WLOG
?

WLOG meams Without Loss of Generality.