secondary regional 2017

For students of class 9-10 (age 14-16)
MD ADNAN RAHMAN
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 $6z-102=34 y$ ...$(1)$ 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.

Tasnood
Posts: 73
Joined: Tue Jan 06, 2015 1:46 pm

Re: secondary regional 2017

Another clear-cut solution
$34x+51y=6z \Rightarrow 34x+51y-6z=0$ Here, $34x$ and, $6z$ are even. So, $34x-6z$ 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: $6z-34x=51 \times 2 \Rightarrow 6z-34x=102 \Rightarrow 3z-17x=51$
$3$ 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

Tasnood wrote:
Thu Feb 15, 2018 11:22 pm
Another clear-cut solution
$34x+51y=6z \Rightarrow 34x+51y-6z=0$ Here, $34x$ and, $6z$ are even. So, $34x-6z$ 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: $6z-34x=51 \times 2 \Rightarrow 6z-34x=102 \Rightarrow 3z-17x=51$
$3$ 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.
Nice solution.Thanks.
MD ADNAN RAHMAN wrote:
Mon Dec 11, 2017 3:29 pm
34x+51y=6z, here x,y,z are prime numbers. find the value of (x=y=z)
I think the actual problem was not like this.The real problem is
Here

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

Re: secondary regional 2017

Hint for the real problem

Tasnood
Posts: 73
Joined: Tue Jan 06, 2015 1:46 pm

Re: secondary regional 2017

$34x+51y=6z \Rightarrow 34x=51y-6z=0$
$34x$ is even for all values of $x$. $6z$ is even too.
Then, $51y$ must be even, $y=2$ then.

We can write: $6z-34x=102$ [Applying the value of $y$]
$\Rightarrow 3z-17x=51$
$3$ divides RS. So, $3$ must divide $(3z-51)$. $3$ divides $3z$. Then $17x$. $3$ doesn't divide $17$. So, $3$ divides $x$.
We write: $x=3a$
Applying the value of $x$, we get: $3z-17 \times 3a=51 \Rightarrow 3z-51a=51 \Rightarrow 3z=51a+51 \Rightarrow z=17+17a \Rightarrow z=17(a+1)$
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

Tasnood wrote:
Fri Feb 16, 2018 6:56 pm
$34x+51y=6z \Rightarrow 34x=51y-6z=0$
$34x$ is even for all values of $x$. $6z$ is even too.
Then, $51y$ must be even, $y=2$ then.

We can write: $6z-34x=102$ [Applying the value of $y$]
$\Rightarrow 3z-17x=51$
$3$ divides RS. So, $3$ must divide $(3z-51)$. $3$ divides $3z$. Then $17x$. $3$ doesn't divide $17$. So, $3$ divides $x$.
We write: $x=3a$
Applying the value of $x$, we get: $3z-17 \times 3a=51 \Rightarrow 3z-51a=51 \Rightarrow 3z=51a+51 \Rightarrow z=17+17a \Rightarrow z=17(a+1)$
If $a \geq 1$, $(a+1) \geq 2$ and, $z$ will be an even number. So, $z=17$.
$a+1=1 \Rightarrow a=0 \Rightarrow 3a=0 \Rightarrow x=0$
$x=0,y=2,z=17$
$x+(y \times z)=0+(2 \times 17)=34$
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.

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 $\fbox {71}$

Tasnood
Posts: 73
Joined: Tue Jan 06, 2015 1:46 pm

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

Re: secondary regional 2017

Tasnood wrote:
Fri Feb 16, 2018 7:45 pm
You didn't write that in the last question.
I get it wrong after posting.So,it is my fault.Sorry