Just for fun!-XVIII

Number Theory Level 3

Given natural numbers $x, y, z$ that satisfy $x+y+z+xy+yz+zx+xyz=1000,$

evaluate $x^{2}+y^{2}+z^{2}.$

The answer is 280.

2 solutions

Swapnil Das
Aug 21, 2015

Given,

$x+y+z+xy+yz+zx+xyz=1000$

$x+y+z+xy+yz+zx+xyz+1=1001$

$(1+x)(1+y)(1+z) = 1001$

Factorizing, $1001 = 7\times 11\times 13$

Let $1+x = 7$ , $1+y=11$ , $1 + z=13$

$x=6$ , $y=10$ , $z=12$

Evaluating the expression gives $6^{2} + 10^{2} + 12^{2} = 280$

Note :

It is very correct if one takes values of $x$ as $10$ and $12$ , and so on for other variables. It won't affect the answer.
Though $1$ is a factor of $1001$ , this value can't be assigned to any other variables as the variables are asked to be natural in the question.

Moderator note:

Yes, adding 1 to both sides of the equation makes the whole problem trivial.

For completeness. You should explain why you only considered $1 + x= 7, 1+y=11, 1 + z = 13$ . Why can't it be $1 + x= 11, 1 + y = 13, 1 + z = 7$ ?

Hint : Without the loss of generality, we can assume $x \leq y \leq z$ .

I did the same ^_^

Akshat Sharda - 5 years, 9 months ago

Just for completeness, you should explain why $x+1$ , $y+1$ and $z+1$ must greater than $1$ ( $x, y, z$ are natural numbers of course :D). Otherwise, your solution is fabulous ^_^

Trung Đặng Đoàn Đức - 5 years, 9 months ago

Thank you for your kind complement. I would bring out necessary changes in the solution very soon.😊

Swapnil Das - 5 years, 9 months ago

very nice....

Dev Sharma - 5 years, 9 months ago

Thank you Sir.

Swapnil Das - 5 years, 9 months ago

no sir..... please

Dev Sharma - 5 years, 9 months ago

are you on fb?

Dev Sharma - 5 years, 9 months ago

@Dev Sharma – No, but I am in G+.

Swapnil Das - 5 years, 9 months ago

@Swapnil Das – ok bro .....

Dev Sharma - 5 years, 9 months ago

@Dev Sharma – Well, why did you ask this question? Do you want to talk to me? I would love to talk to.

Swapnil Das - 5 years, 9 months ago

@Swapnil Das – yes, i want to ask you something

Dev Sharma - 5 years, 9 months ago

@Dev Sharma – Feel free to do so.

Swapnil Das - 5 years, 9 months ago

@Swapnil Das – are you preparing for olympiad?

Dev Sharma - 5 years, 9 months ago

@Dev Sharma – Yeah, I am. Only Mathematics Olympiads. I hate to do for science as I am not good at Chem and Bio.

Swapnil Das - 5 years, 9 months ago

@Swapnil Das – Which books do you use?

Dev Sharma - 5 years, 9 months ago

@Dev Sharma – I use 11 class RD Sharma for concepts, RMO and Inmo by Rajeev Manocha and also Hall and knight Algebra.

Swapnil Das - 5 years, 9 months ago

@Swapnil Das – and Brilliant too.

Dev Sharma - 5 years, 9 months ago

@Dev Sharma – Ya, its the thing which changed my life and made me learn Calculus.

Swapnil Das - 5 years, 9 months ago

Good Job! Since $x,y,z$ are symmetric, without loss of generality, you can assume from the start that $x \leq y \leq z$ , to complete the solution.

Satyajit Mohanty - 5 years, 9 months ago

Can we say that x>y>z , Could you please add some more explanation for beginners as I am thinking different combination.

Syed Baqir - 5 years, 9 months ago

Yes, you can also assume without loss of generality that $x \geq y \geq z$ or you can also assume $y \geq x \geq z$ , any combination, doesn't matter, but once you assumed something, you have to solve the problem with that particular assumption only.

Since the expression in the problem statement, and the value of expression asked, both are symmetric in nature, any of the assumptions doesn't change the final answer, thereby.

Satyajit Mohanty - 5 years, 9 months ago

@Satyajit Mohanty – Satyajit Bhai, Can you please come to Slack? I want to ask you many things :P

Swapnil Das - 5 years, 9 months ago

@Swapnil Das – For information to non-Indians, the term "Bhai" in India means "Brother" :D

@Swapnil Das Sure!

Satyajit Mohanty - 5 years, 9 months ago

@Swapnil Das – sorry to disturb, but whats slack?

Dev Sharma - 5 years, 9 months ago

@Dev Sharma – It's a platform for Brilliant Users to chat with each other on various topics, mostly used for developing Brilliant and its community standards.

Link to Brilliant Lounge on Slack

Satyajit Mohanty - 5 years, 9 months ago

@Dev Sharma – A place where we Brilliantians talk. You may also join.

Swapnil Das - 5 years, 9 months ago

@Swapnil Das – How to join?

Dev Sharma - 5 years, 9 months ago

@Dev Sharma – Link to Brilliant Lounge on Slack

Visit this link. You'll automatically know what to do next to join.

Satyajit Mohanty - 5 years, 9 months ago

@Satyajit Mohanty – very nice explanation, thank you

Syed Baqir - 5 years, 9 months ago

Can you add some more lines how the second line is transformed to third line:

Because I am thinking like this and I am stuck ,

x(1+y) + y(1+z) + z(1+x) + xyz + 1 = 1001

Syed Baqir - 5 years, 9 months ago

$x + y + z + x\cdot y + y\cdot z + z\cdot x + x\cdot y\cdot z +1$

$= (z +1) + x + y + x\cdot y + y\cdot z + z\cdot x + x\cdot y\cdot z$

$= (z +1) + x\cdot (y + 1) + y + y\cdot z + z\cdot x + x\cdot y\cdot z$

$= (z +1) + x\cdot (y + 1) + y\cdot (z + 1) + z\cdot x + x\cdot y\cdot z$

$= (z +1) + x\cdot (y + 1) + y\cdot (z + 1) + z\cdot x\cdot (y + 1)$

$= (z +1)\cdot (y + 1) + x\cdot (y + 1) + z\cdot x\cdot (y + 1)$

$= (y + 1)\cdot \Big((z + 1) + x + z\cdot x\Big)$

$= (y + 1)\cdot \Big((z + 1) + x\cdot (z + 1)\Big)$

$= (y + 1)\cdot \Big((z + 1)\cdot (x + 1)\Big)$

$\boxed{= (y + 1)\cdot (z + 1)\cdot (x + 1)}$

Hope this helps kuprie ;)

Chris Galanis - 5 years, 9 months ago

Aakash Khandelwal
Aug 22, 2015

Add 1 and question is yours

I see that you always post these kinds of comments in the solutions section. I think it'll be good if you post these comments in the comments section of any solution. If you want to post any solution, post the complete one.

Satyajit Mohanty - 5 years, 9 months ago

Just for fun!-XVIII

The answer is 280.

2 solutions

Moderator note:

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