BDMO National Junior 2016/6

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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
 Posts: 86
 Joined: Thu Aug 20, 2015 7:11 pm
 Location: Malibagh,Dhaka1217
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
 Location: Rajshahi,Bangladesh
Re: BDMO National Junior 2016/6
But how?....Please explain it.
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, $3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$
$\Rightarrow 3^{2w} ( 1 + 3^{3x2w} + 3^{5y2w}) = 3^{7z}$
$\Rightarrow 1 + 3^{3x2w} + 3^{5y2w} = 3^{7z2w}$
Which gives that R.H.S is divisible by $3$, but L.H.S is not unless $3^{3x2w} = 3^{5y3x} = 1.$ 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 ,
$\Rightarrow 1 + 1 + 3^{5y2w} = 3^{7z2w}$.
But here also, R.H.S is divisible by $3$, but the L.H.S is not unless $3^{5y2w} = 1$, 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, $w + x + y + z = 13 + 45 + 30 + 18 = 106$
WLOG, Let's assume, $3^{2w} < 3^{3x} < 3^{5y}$
Then, $3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$
$\Rightarrow 3^{2w} ( 1 + 3^{3x2w} + 3^{5y2w}) = 3^{7z}$
$\Rightarrow 1 + 3^{3x2w} + 3^{5y2w} = 3^{7z2w}$
Which gives that R.H.S is divisible by $3$, but L.H.S is not unless $3^{3x2w} = 3^{5y3x} = 1.$ 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 ,
$\Rightarrow 1 + 1 + 3^{5y2w} = 3^{7z2w}$.
But here also, R.H.S is divisible by $3$, but the L.H.S is not unless $3^{5y2w} = 1$, 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
 Charles Caleb Colton
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 $3^{2w} = 3^{3x} = 3^{5y} = 3^{7z1}$
Or, $2w=3x=5y=7z1$
So. $w+x+y+z=45+30+18+13=106$
$3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$
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 $3^{2w} = 3^{3x} = 3^{5y} = 3^{7z1}$
Or, $2w=3x=5y=7z1$
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.
 ahmedittihad
 Posts: 181
 Joined: Mon Mar 28, 2016 6:21 pm
Re: BDMO National Junior 2016/6
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 wrote:$3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$
WLOG, Let's assume, $3^{2w} < 3^{3x} < 3^{5y}$
Then, $3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$
$\Rightarrow 3^{2w} ( 1 + 3^{3x2w} + 3^{5y2w}) = 3^{7z}$
$\Rightarrow 1 + 3^{3x2w} + 3^{5y2w} = 3^{7z2w}$
Which gives that R.H.S is divisible by $3$, but L.H.S is not unless $3^{3x2w} = 3^{5y3x} = 1.$ 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 ,
$\Rightarrow 1 + 1 + 3^{5y2w} = 3^{7z2w}$.
But here also, R.H.S is divisible by $3$, but the L.H.S is not unless $3^{5y2w} = 1$, 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, $w + x + y + z = 13 + 45 + 30 + 18 = 106$
Frankly, my dear, I don't give a damn.
Re: BDMO National Junior 2016/6
$3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$
Let's assume, $3^{2w} < 3^{3x} < 3^{5y}$
Then, $3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$
$\Rightarrow 3^{2w} ( 1 + 3^{3x2w} + 3^{5y2w}) = 3^{7z}$
$\Rightarrow 1 + 3^{3x2w} + 3^{5y2w} = 3^{7z2w}$
Which gives that R.H.S is divisible by $3$, but L.H.S is not unless $3^{3x2w} = 3^{5y3x} = 1.$ 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 ,
$\Rightarrow 1 + 1 + 3^{5y2w} = 3^{7z2w}$.
But here also, R.H.S is divisible by $3$, but the L.H.S is not unless $3^{5y2w} = 1$, 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.
Let's assume, $3^{2w} < 3^{3x} < 3^{5y}$
Then, $3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$
$\Rightarrow 3^{2w} ( 1 + 3^{3x2w} + 3^{5y2w}) = 3^{7z}$
$\Rightarrow 1 + 3^{3x2w} + 3^{5y2w} = 3^{7z2w}$
Which gives that R.H.S is divisible by $3$, but L.H.S is not unless $3^{3x2w} = 3^{5y3x} = 1.$ 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 ,
$\Rightarrow 1 + 1 + 3^{5y2w} = 3^{7z2w}$.
But here also, R.H.S is divisible by $3$, but the L.H.S is not unless $3^{5y2w} = 1$, 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
 Charles Caleb Colton
 Thanic Nur Samin
 Posts: 176
 Joined: Sun Dec 01, 2013 11:02 am
Re: BDMO National Junior 2016/6
All you did was drop the word WLOG. You can't just assume that.dshasan wrote: Let's assume, $3^{2w} < 3^{3x} < 3^{5y}$
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.
Because destroying everything mindlessly isn't cool enough.
 samiul_samin
 Posts: 1007
 Joined: Sat Dec 09, 2017 1:32 pm