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

BDMO National Junior 2016/6

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BDMO National Junior 2016/6

Post Number:#1  Unread postby Math Mad Muggle » Wed Jan 25, 2017 12:30 am

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)
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Re: BDMO National Junior 2016/6

Post Number:#2  Unread postby Kazi_Zareer » Wed Jan 25, 2017 1:58 am

$w+x+y+z=106$
We cannot solve our problems with the same thinking we used when we create them.
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Re: BDMO National Junior 2016/6

Post Number:#3  Unread postby Math Mad Muggle » Wed Jan 25, 2017 8:25 am

But how?....Please explain it.
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Re: BDMO National Junior 2016/6

Post Number:#4  Unread postby dshasan » Wed Jan 25, 2017 2:00 pm

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

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Re: BDMO National Junior 2016/6

Post Number:#5  Unread postby Akhiar » Thu Jan 26, 2017 7:45 pm

What do yo mean by WLOG
?
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Re: BDMO National Junior 2016/6

Post Number:#6  Unread postby Thamim Zahin » Tue Jan 31, 2017 8:11 pm

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. $w+x+y+z=45+30+18+13=106$
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Re: BDMO National Junior 2016/6

Post Number:#7  Unread postby ahmedittihad » Wed Feb 01, 2017 10:25 am

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.
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Re: BDMO National Junior 2016/6

Post Number:#8  Unread postby dshasan » Thu Feb 02, 2017 11:52 am

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

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.

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Re: BDMO National Junior 2016/6

Post Number:#9  Unread postby Thanic Nur Samin » Thu Feb 02, 2017 2:12 pm

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 $\alpha < \beta < \gamma$.
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