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# Olympiad Practice Problems Part 1

Some problems from NMTC level 2 junior contest, conducted in India by the AMTI. Try to solve 'em all! And please post solutions, like, and re-share!

1 [NMTC 2011]

If $${a=2011}^{2010}$$, $$b={2010}^{2011}$$, $$c={4021}^{4021}$$, and $$d=2011$$, find the value of $\frac{bc(a+d)}{(a-b)(a-c)} + \frac{ac(b+d)}{(b-a)(b-c)} + \frac{ab(a+d)}{(c-a)(c-b)}$

2 [NMTC 2011]

$$x$$, $$y$$, $$z$$ and $$a$$ are real numbers such that $$x+y+z=3$$ and $$xy+yz+zx=a$$. The difference in maximum and minimum possible value of $$x$$ is $$8$$. Find $$a$$.

3 [NMTC 2012]

Find all positive integral solutions $$(x,y)$$ to the equation $$4x^3-3x-1=2y^2$$

4 [NMTC 2012]

Consider the set of numbers $$A={1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, ........, \frac{1}{2012}}$$. We delete any two numbers $$a$$ and $$b$$ from the set $$A$$, and replace them with a single number $$ab+a+b$$. What number will be left at the end, after $$2011$$ such operations. Will the number always be the same?

5 [NMTC 2013]

Find positive real solutions $$(x,y,z)$$ which satisfy both these equations: $$x^3+y^3+z^3=x+y+z$$ and $$x^2+y^2+z^2=xyz$$.

6 [NMTC 2013]

Do there exis $$10$$ distinct integers, such that the sum of any nine of them is a perfect square?

Note by Satvik Golechha
3 years, 2 months ago

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for 4 take product of each number increased by 1 as invariant initial value is 2013 so answer is at final step 2013-1=2012

- 3 years, 2 months ago

- 3 years, 2 months ago

Consider product (a+1)*(b+1) when a and b are deleted and ab+a+b is added but it is equal to product.so this will remain constant at all steps?hope u got it

- 3 years, 2 months ago

There exist an infinite number of solutions for equation in problem nr.3,given by:

$$x=2a^{2}+1$$

$$y=a(4a^{2}+3)$$ where $$a$$ can be any positive integer.

- 3 years, 2 months ago

Here's how i came up with that:

$$3x^{3}-3x+x^{3}-1=2y^{2}$$.

$$=3x(x^{2}-1)+(x-1)(x^{2}+x+1)=2y^2$$

$$=(4x^{2}+4x+1)(x-1)=2y^2$$.

$$=\frac{(2x+1)^{2}(x-1)}{2}=y^{2}$$. Since $$(2x+1)^{2}$$ is always odd,then $$x-1$$ must be divisible by $$2$$ or $$x$$ must be an odd number.Also we have the product of two numbers where one of them is a square and that the whole product is a square,thus $$\frac{x-1}{2}$$ must be a perfect square therefore $$x=2a^{2}+1$$ where $$a^2=\frac{x-1}{2}$$.Substituting $$x$$ we have $$y=a(4a^2+3)$$.

- 3 years, 2 months ago

Looks great!

Staff - 3 years, 2 months ago

Brilliant! *That is the only thing I have to say.

- 3 years, 2 months ago

Yes, that is a family of solutions.

Is that the only family? Can there be other solutions which are not of that form?

Staff - 3 years, 2 months ago

No,because the way this is derived shows that there is no room for other family of solutions.

- 3 years, 2 months ago

No, I have a proof of that

- 3 years, 2 months ago

For the second one, I got $$\boxed{a=-9}$$

Take the second equation and differentiate it with respect to $$x$$, to get $$f'(x)=y+z$$. Extrema is reached when this is equal to zero, and as there is no constraint on interval, $$y+z=0$$. Plugging into the first equation, we get $$x=3$$. This must be the minimum, since the other extreme of $$x$$ is $$5$$, which is the maximum. For the maximum, $$y+z=-2$$ or $$z=-y-2$$

Now, manipulating the second equation,

$xy+z(x+y)=a$

$xy-(y+2)(x+y)=a$

$-(y^2+2y+10)=a$

Completing the square, we get $$(y+1)^2+9=-a$$

Plugging in $$z$$, we get $$(z+1)^2+9=-a$$

Here my approach was a bit unfounded.

I found the minimum value of $$a$$ in both expressions, which is achieved when $$(y+1)^2,(z+1)^2=0$$

Thus, $$x=5, y=-1, z=-1, \boxed{a=-9}$$, is a solution set where $$x$$ is maximized.

- 3 years, 2 months ago

Hey @Nanayaranaraknas Vahdam how do you know calculus at 15 i.e in 10th class, in rajasthan the study pattern's diffrerent. But what abt. there please tell me?????Please!(But i know only Minima and Maxima)

- 3 years, 2 months ago

I didn't learn it through school, I did some independent studying, and learned some basics in Calculus.

P.S. I am in 11th

- 3 years, 2 months ago

Oh u are in 11th ,,,actually i am in 10th

- 3 years, 2 months ago

This HAS to be solved without Calculus-So this method would be better. $$y+z=3-x$$ and $$yz=a-x(3-x)$$. Manipulate these , form quadratics, use the discriminant and you're done. :D

- 3 years, 2 months ago

I have done it like:

$${x}^{2} + {y}^{2} + {z}^{2} = 9 - 2a$$

$${y}^{2} + {z}^{2} = {(3-x)}^{2} - 2(a-x(3-x))$$

$$9-2a - {x}^{2} = {(3-x)}^{2} - 2(a-x(3-x))$$

This is the approach you are talking about?

- 3 years, 2 months ago

What is the answer you arrived at?

- 3 years, 2 months ago

$$-9$$ only

- 3 years, 2 months ago

thank you very much!!!

- 3 years, 2 months ago

Try the second part too!

- 3 years, 2 months ago

try a angular kinematics problem from me . rotating ring.

- 3 years, 2 months ago

the third one , (x-1)(2x+1)^2=2y^2 ,y=(2x+1)sqrt{(x-1)}/2 for y to be an integer (x-1)/2=m^2 ,x=2*m^2+1 then for m=0,1,2,3,4,5........ we have infinite number of solutions

- 3 years, 2 months ago

In 5 note x y z all can't be greater than 1. Due to first equation if all are 1 it is one tulle is pf any of them is less than 1 say x then x^2+y^2+z^2>y^2+z^2>yz so. Second fails so only 1,1,1 is solution

- 3 years, 2 months ago

Another approach maybe $${x}^{3} + {y}^{3} + {z}^{3} \geq 3xyz$$

But, this - $$x + y + z \geq 3({x}^{2} + {y}^{2} + {z}^{2})$$ for positive integer (x,y,z) as $$3({x}^{2} + {y}^{2} + {z}^{2})$$ will increase more than $$x + y + z$$ for positive integers(x,y,z)

- 3 years, 2 months ago

yes by cauchy scwarz inequality (x+y+z)^2<=3(x^2+y^2+z^2).so your method is also elegant

- 3 years, 2 months ago

sorry for the mistake. so in case 1,1,1 too no solution exist hence there is no solution to 5.

- 3 years, 2 months ago

5actually doesn't possess a solution

- 3 years, 2 months ago

But $${1}^{2} +{1}^{2} + {1}^{2} \neq 1*1*1$$

- 3 years, 2 months ago

Exactly.

- 3 years, 2 months ago

After bashing a lot, I have simplified the 1st to $$d - \frac{ab}{c-b}$$. Is it correct? I am really not sure about it.

- 3 years, 2 months ago

I think this much simplification would do. It's value is 2010.99999999999999....., so they will accept 2011.

- 3 years, 2 months ago

Yeah! Absolutely! I was about to write that $$\frac{ab}{c-b}$$ is almost equal to 1.

BTW, @Satvik Golechha are you planning to make more such sets like for NT or geometry?

- 3 years, 2 months ago

@Kartik Sharma Did you like it? These were some of my doubts. And yes, if people like it, I would definitely continue. Geometry is my Achilles heel, I need to study.

- 3 years, 2 months ago

That expression is algebraically correct.

How would you proceed to do the calculation of the final value?

Staff - 3 years, 2 months ago

For #4)-The second part of the question was not tested in the paper. $$2012$$ as the answer is easily guess-able but quite hard to prove.

- 3 years, 2 months ago

Will you please prove it @Krishna Ar ? Thanks.

- 3 years, 2 months ago

^6) Is already discussed on Brilliant.

- 3 years, 2 months ago

Where?

- 3 years, 2 months ago

Comment deleted Oct 28, 2014

It is never said that the integer are positive

- 3 years, 2 months ago

Why must the 10 distinct positive integers have last digits of $$0, 1, 2, 3, 4, 5, 6, 7, 8, 9$$?

Why can't the 10 distinct positive integers all have a last digit of 1?

Note: I believe that the answer is Yes.

Hint: Find the sum of 10 distinct squares which is the multiple of 9.

Staff - 3 years, 2 months ago

Oh yes, thanks. I mistook the question.

- 3 years, 2 months ago

Yes ,you are right.

- 3 years, 2 months ago

did the same!

- 3 years, 2 months ago