secondary regional 2017

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MD ADNAN RAHMAN
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secondary regional 2017

Unread post by MD ADNAN RAHMAN » Mon Dec 11, 2017 3:29 pm

34x+51y=6z, here x,y,z are prime numbers. find the value of (x=y=z)

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samiul_samin
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Re: secondary regional 2017

Unread post by samiul_samin » Wed Feb 14, 2018 12:02 am

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.

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Tasnood
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Re: secondary regional 2017

Unread post by Tasnood » 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. :P

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samiul_samin
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Re: secondary regional 2017

Unread post by samiul_samin » Thu Feb 15, 2018 11:36 pm

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. :P
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
($x,y,z$) are three non negative integers that satisfies $34x+51y=6z$, if $y$ and $z$ are primes,then what is the value of $x+(y×z)$ ?

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samiul_samin
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Re: secondary regional 2017

Unread post by samiul_samin » Fri Feb 16, 2018 8:08 am

Hint for the real problem
2 is the only even prime.

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Tasnood
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Re: secondary regional 2017

Unread post by Tasnood » 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, $z=17$.
$a+1=1 \Rightarrow a=0 \Rightarrow 3a=0 \Rightarrow x=0$
$x=0,y=2,z=17$
$x+(y \times z)=0+(2 \times 17)=34$ :?:

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samiul_samin
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Re: secondary regional 2017

Unread post by samiul_samin » Fri Feb 16, 2018 7:17 pm

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.

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samiul_samin
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Re: secondary regional 2017

Unread post by samiul_samin » Fri Feb 16, 2018 7:21 pm

According to my last post $(x,y,z)=(3,2,34) $ so, the answer is $\fbox {71}$

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Tasnood
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Re: secondary regional 2017

Unread post by Tasnood » Fri Feb 16, 2018 7:45 pm

You didn't write that in the last question. :?:

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samiul_samin
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Re: secondary regional 2017

Unread post by samiul_samin » Fri Feb 16, 2018 8:05 pm

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 :cry: :cry: :cry:

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