## BDMO National Junior 2016/6

For students of class 6-8 (age 12 to 14)
Posts: 29
Joined: Mon Jan 23, 2017 10:32 am

### BDMO National Junior 2016/6

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
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Location: Malibagh,Dhaka-1217

### Re: BDMO National Junior 2016/6

We cannot solve our problems with the same thinking we used when we create them.

Posts: 29
Joined: Mon Jan 23, 2017 10:32 am

### Re: BDMO National Junior 2016/6

dshasan
Posts: 66
Joined: Fri Aug 14, 2015 6:32 pm

### Re: BDMO National Junior 2016/6

$3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$
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, $w + x + y + z = 13 + 45 + 30 + 18 = 106$
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.

- Charles Caleb Colton

Akhiar
Posts: 1
Joined: Tue Oct 11, 2016 12:39 am

### Re: BDMO National Junior 2016/6

What do yo mean by WLOG
?

Thamim Zahin
Posts: 98
Joined: Wed Aug 03, 2016 5:42 pm

### Re: BDMO National Junior 2016/6

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. $w+x+y+z=45+30+18+13=106$
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.

Posts: 181
Joined: Mon Mar 28, 2016 6:21 pm

### Re: BDMO National Junior 2016/6

dshasan wrote:$3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$
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.
Frankly, my dear, I don't give a damn.

dshasan
Posts: 66
Joined: Fri Aug 14, 2015 6:32 pm

### Re: BDMO National Junior 2016/6

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

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. The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.

- Charles Caleb Colton

Thanic Nur Samin
Posts: 176
Joined: Sun Dec 01, 2013 11:02 am

### Re: BDMO National Junior 2016/6

dshasan wrote: 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 $\alpha < \beta < \gamma$.
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

samiul_samin
Posts: 1007
Joined: Sat Dec 09, 2017 1:32 pm

### Re: BDMO National Junior 2016/6

Akhiar wrote:
Thu Jan 26, 2017 7:45 pm
What do yo mean by WLOG
?

WLOG
meams Without Loss of Generality.