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BdMO Online Forum • View topic - BDMO National Junior 2016/6

## BDMO National Junior 2016/6

For students of class 6-8 (age 12 to 14)

### 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)

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

### Re: BDMO National Junior 2016/6

$w+x+y+z=106$
We cannot solve our problems with the same thinking we used when we create them.

Kazi_Zareer

Posts: 86
Joined: Thu Aug 20, 2015 7:11 pm
Location: Malibagh,Dhaka-1217

### Re: BDMO National Junior 2016/6

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

### 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
dshasan

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

### Re: BDMO National Junior 2016/6

What do yo mean by WLOG
?
Akhiar

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

### 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.

Thamim Zahin

Posts: 98
Joined: Wed Aug 03, 2016 5:42 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.

Posts: 147
Joined: Mon Mar 28, 2016 6:21 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
dshasan

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

### 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.

Thanic Nur Samin

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