secondary regional 2017

 Posts: 3
 Joined: Sun Dec 10, 2017 3:59 pm
secondary regional 2017
34x+51y=6z, here x,y,z are prime numbers. find the value of (x=y=z)
 samiul_samin
 Posts: 1007
 Joined: Sat Dec 09, 2017 1:32 pm
Re: secondary regional 2017
There is no such $x,y,z$ that all of them are prime.
According to the question the right side of the equation is even. So the sum of the two terms of left side should be even. As $34x$ is even $51y $ should be even.so $y=2$. Now we can say get that $6z102=34 y$ ...$(1)$ if $y=3$ then $z$ is not a prime.
If $y>3 $ and $y$ is a prime then $y$ don't have 3 in its PPF.
So, the right side of the $(1)$ equation is not divided by 6.But the left side is a multiple of 6.
A CONTRADICTION and We are done.
According to the question the right side of the equation is even. So the sum of the two terms of left side should be even. As $34x$ is even $51y $ should be even.so $y=2$. Now we can say get that $6z102=34 y$ ...$(1)$ if $y=3$ then $z$ is not a prime.
If $y>3 $ and $y$ is a prime then $y$ don't have 3 in its PPF.
So, the right side of the $(1)$ equation is not divided by 6.But the left side is a multiple of 6.
A CONTRADICTION and We are done.
Re: secondary regional 2017
Another clearcut solution
$34x+51y=6z \Rightarrow 34x+51y6z=0$ Here, $34x$ and, $6z$ are even. So, $34x6z$ is even. As LS is even $(0)$, $51y$ must be even. $51$ is not even. So, $y$ is obviously even and, $y=2$
We can write: $6z34x=51 \times 2 \Rightarrow 6z34x=102 \Rightarrow 3z17x=51$
$3$ divides LS. So, $3$ must divide RS too.
$3$ divides $3z$. So, $3$ must divide $17x$. $3$ doesn't divide $17$, then $3$ must divide $x$ and, $x=3$
Applying $y=2,x=3$, we get:
$6z=34 \times 3+51 \times 2=204 \Rightarrow z=34$
But, $34$ is not prime
So, the value of $(x,y,z)$ is invalid.
$34x+51y=6z \Rightarrow 34x+51y6z=0$ Here, $34x$ and, $6z$ are even. So, $34x6z$ is even. As LS is even $(0)$, $51y$ must be even. $51$ is not even. So, $y$ is obviously even and, $y=2$
We can write: $6z34x=51 \times 2 \Rightarrow 6z34x=102 \Rightarrow 3z17x=51$
$3$ divides LS. So, $3$ must divide RS too.
$3$ divides $3z$. So, $3$ must divide $17x$. $3$ doesn't divide $17$, then $3$ must divide $x$ and, $x=3$
Applying $y=2,x=3$, we get:
$6z=34 \times 3+51 \times 2=204 \Rightarrow z=34$
But, $34$ is not prime
So, the value of $(x,y,z)$ is invalid.
 samiul_samin
 Posts: 1007
 Joined: Sat Dec 09, 2017 1:32 pm
Re: secondary regional 2017
Nice solution.Thanks.Tasnood wrote: ↑Thu Feb 15, 2018 11:22 pmAnother clearcut solution
$34x+51y=6z \Rightarrow 34x+51y6z=0$ Here, $34x$ and, $6z$ are even. So, $34x6z$ is even. As LS is even $(0)$, $51y$ must be even. $51$ is not even. So, $y$ is obviously even and, $y=2$
We can write: $6z34x=51 \times 2 \Rightarrow 6z34x=102 \Rightarrow 3z17x=51$
$3$ divides LS. So, $3$ must divide RS too.
$3$ divides $3z$. So, $3$ must divide $17x$. $3$ doesn't divide $17$, then $3$ must divide $x$ and, $x=3$
Applying $y=2,x=3$, we get:
$6z=34 \times 3+51 \times 2=204 \Rightarrow z=34$
But, $34$ is not prime
So, the value of $(x,y,z)$ is invalid.
I think the actual problem was not like this.The real problem isMD ADNAN RAHMAN wrote: ↑Mon Dec 11, 2017 3:29 pm34x+51y=6z, here x,y,z are prime numbers. find the value of (x=y=z)
Here
 samiul_samin
 Posts: 1007
 Joined: Sat Dec 09, 2017 1:32 pm
Re: secondary regional 2017
Hint for the real problem
Re: secondary regional 2017
$34x+51y=6z \Rightarrow 34x=51y6z=0$
$34x$ is even for all values of $x$. $6z$ is even too.
Then, $51y$ must be even, $y=2$ then.
We can write: $6z34x=102$ [Applying the value of $y$]
$ \Rightarrow 3z17x=51$
$3$ divides RS. So, $3$ must divide $(3z51)$. $3$ divides $3z$. Then $17x$. $3$ doesn't divide $17$. So, $3$ divides $x$.
We write: $x=3a$
Applying the value of $x$, we get: $3z17 \times 3a=51 \Rightarrow 3z51a=51 \Rightarrow 3z=51a+51 \Rightarrow z=17+17a \Rightarrow z=17(a+1)$
If $a \geq 1$, $(a+1) \geq 2$ and, $z$ will be an even number. So, $z=17$.
$a+1=1 \Rightarrow a=0 \Rightarrow 3a=0 \Rightarrow x=0$
$x=0,y=2,z=17$
$x+(y \times z)=0+(2 \times 17)=34$
$34x$ is even for all values of $x$. $6z$ is even too.
Then, $51y$ must be even, $y=2$ then.
We can write: $6z34x=102$ [Applying the value of $y$]
$ \Rightarrow 3z17x=51$
$3$ divides RS. So, $3$ must divide $(3z51)$. $3$ divides $3z$. Then $17x$. $3$ doesn't divide $17$. So, $3$ divides $x$.
We write: $x=3a$
Applying the value of $x$, we get: $3z17 \times 3a=51 \Rightarrow 3z51a=51 \Rightarrow 3z=51a+51 \Rightarrow z=17+17a \Rightarrow z=17(a+1)$
If $a \geq 1$, $(a+1) \geq 2$ and, $z$ will be an even number. So, $z=17$.
$a+1=1 \Rightarrow a=0 \Rightarrow 3a=0 \Rightarrow x=0$
$x=0,y=2,z=17$
$x+(y \times z)=0+(2 \times 17)=34$
 samiul_samin
 Posts: 1007
 Joined: Sat Dec 09, 2017 1:32 pm
Re: secondary regional 2017
Because $x,y,z$ all are greater than 0.So, the question will be that {$x,y$ are prime and $z$ is composite} The real question in the exam was wrong.Tasnood wrote: ↑Fri Feb 16, 2018 6:56 pm$34x+51y=6z \Rightarrow 34x=51y6z=0$
$34x$ is even for all values of $x$. $6z$ is even too.
Then, $51y$ must be even, $y=2$ then.
We can write: $6z34x=102$ [Applying the value of $y$]
$ \Rightarrow 3z17x=51$
$3$ divides RS. So, $3$ must divide $(3z51)$. $3$ divides $3z$. Then $17x$. $3$ doesn't divide $17$. So, $3$ divides $x$.
We write: $x=3a$
Applying the value of $x$, we get: $3z17 \times 3a=51 \Rightarrow 3z51a=51 \Rightarrow 3z=51a+51 \Rightarrow z=17+17a \Rightarrow z=17(a+1)$
If $a \geq 1$, $(a+1) \geq 2$ and, $z$ will be an even number. So, $z=17$.
$a+1=1 \Rightarrow a=0 \Rightarrow 3a=0 \Rightarrow x=0$
$x=0,y=2,z=17$
$x+(y \times z)=0+(2 \times 17)=34$
 samiul_samin
 Posts: 1007
 Joined: Sat Dec 09, 2017 1:32 pm
Re: secondary regional 2017
According to my last post $(x,y,z)=(3,2,34) $ so, the answer is $\fbox {71}$
Re: secondary regional 2017
You didn't write that in the last question.
 samiul_samin
 Posts: 1007
 Joined: Sat Dec 09, 2017 1:32 pm