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?

## Comments

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TopNewestfor 4 take product of each number increased by 1 as invariant initial value is 2013 so answer is at final step 2013-1=2012 – Gourav Bhattad · 2 years ago

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– Satvik Golechha · 2 years ago

Please explain. Thanks... :DLog in to reply

– Gourav Bhattad · 2 years 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 itLog in to reply

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. – Lawrence Bush · 2 years ago

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

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

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

\(=\frac{(2x+1)^{2}(x-1)}{2}=y^{2}\). 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 \(y=a(4a^2+3)\). – Lawrence Bush · 2 years ago

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– Calvin Lin Staff · 2 years ago

Looks great!Log in to reply

– Kartik Sharma · 2 years ago

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

Is that the only family? Can there be other solutions which are not of that form? – Calvin Lin Staff · 2 years ago

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– Lawrence Bush · 2 years ago

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

– Subrata Saha · 2 years ago

No, I have a proof of thatLog in to reply

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. – Nanayaranaraknas Vahdam · 2 years ago

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@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) – Mehul Chaturvedi · 2 years ago

HeyLog in to reply

P.S. I am in 11th – Nanayaranaraknas Vahdam · 2 years ago

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– Mehul Chaturvedi · 2 years ago

Oh u are in 11th ,,,actually i am in 10thLog in to reply

– Krishna Ar · 2 years 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. :DLog in to reply

\({x}^{2} + {y}^{2} + {z}^{2} = 9 - 2a\)

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

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

This is the approach you are talking about? – Kartik Sharma · 2 years ago

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– Nanayaranaraknas Vahdam · 2 years ago

What is the answer you arrived at?Log in to reply

– Krishna Ar · 2 years ago

\(-9\) onlyLog in to reply

thank you very much!!! – Gaurav Jain · 1 year, 11 months ago

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– Satvik Golechha · 1 year, 11 months ago

Try the second part too!Log in to reply

– Gaurav Jain · 1 year, 11 months ago

try a angular kinematics problem from me . rotating ring.Log in to reply

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 – Ahmed Abdo · 1 year, 11 months agoLog in to reply

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 – Gourav Bhattad · 2 years ago

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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) – Kartik Sharma · 2 years ago

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– Gourav Bhattad · 2 years ago

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

– Gourav Bhattad · 2 years ago

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

– Subrata Saha · 2 years ago

5actually doesn't possess a solutionLog in to reply

– Kartik Sharma · 2 years ago

But \({1}^{2} +{1}^{2} + {1}^{2} \neq 1*1*1 \)Log in to reply

– Satvik Golechha · 2 years ago

Exactly.Log in to reply

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. – Kartik Sharma · 2 years ago

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– Satvik Golechha · 2 years ago

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

BTW, @Satvik Golechha are you planning to make more such sets like for NT or geometry? – Kartik Sharma · 2 years ago

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@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. – Satvik Golechha · 2 years ago

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How would you proceed to do the calculation of the final value? – Calvin Lin Staff · 2 years ago

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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. – Krishna Ar · 2 years ago

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@Krishna Ar ? Thanks. – Satvik Golechha · 2 years ago

Will you please prove itLog in to reply

^6) Is already discussed on Brilliant. – Krishna Ar · 2 years ago

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– Satvik Golechha · 2 years ago

Where?Log in to reply

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– Krishna Ar · 2 years ago

It is never said that the integer are positiveLog in to reply

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. – Calvin Lin Staff · 2 years ago

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– Kartik Sharma · 2 years ago

Oh yes, thanks. I mistook the question.Log in to reply

– Lawrence Bush · 2 years ago

Yes ,you are right.Log in to reply

– Kartik Sharma · 2 years ago

did the same!Log in to reply