1.(a). If \(a,b,c\) are positive reals and \(a+b+c=50\) and \(3a+b-c=70\). If \(x=5a+4b+2c\), find the range of the values of \(x\).

(b). The sides \(a,b,c\) of \(\Delta ABC\) satisfy the equation : \[a^2+2b^2+2016c^2-3ab-4033bc+2017ac=0\]

Prove that \(b\) is the arithmetic mean of \(a,c\).

2.In an isosceles \(\Delta ABC\), \(AB=AC\). The bisector \(AD\) of \(\angle A\) meets the side \(BC\) at \(D\). The line perpendicular to \(AD\) through \(D\) meets \(AB\) at \(F\) and \(AC\) produced at \(E\). Perpendiculars from \(B\) and \(D\) to \(AC\) are \(BM\) and \(DN\) respectively. If \(AE=2016\) units, find the length of \(MN\).

3.(a). Two circles with centres \(P\) and \(Q\) and radii \(\sqrt2\) and \(1\) respectively intersect each other at \(A\) and \(D\) and \(PQ=2\) units. Chord \(AC\) is drawn to the bigger circle to cut it at \(C\) and the smaller circle at \(B\) such that \(B\) is the midpoint of \(AC\). Find the length of \(AC\).

(b). Find the greatest common divisor of the numbers \(n^2-n\) where \(n=3,5,7,9,\ldots\)

4.(a). A book contained problems on Algebra, Geometry and Number Theory. Mahadevan solved some of them. After checking the answers, we found that he answered correctly \(50 \%\) problems in Algebra, \(70 \%\) in Geometry and \(80 \%\) in Number Theory. He further found that he solved correctly \(62 \%\) of problems in Algebra and Number Theory put together, \(74 \%\) questions in Geometry and Number Theory put together. What is the percentage of correctly answered questions in all the three subjects?

(b). Find all pairs of positive integers \((a,b)\) such that \(a^b-b^a=3\).

5.\(a,b,c\) are positive real numbers. Find the minimum value of \[\dfrac{a+3c}{a+2b+c}+\dfrac{4b}{a+b+2c}-\dfrac{8c}{a+b+3c}\]

6.(a). Show that among any \(n+1\) whole numbers, one can find two numbers such that their difference is divisible by \(n\).

(b). Show that for any natural number \(n\), there is a positive integer all of whose digits are \(5\) or \(0\) and is divisible by \(n\).

## Comments

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TopNewestProblem 1b

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– Svatejas Shivakumar · 9 months, 4 weeks ago

I think that you will some marks because you worked backwards(assumed that what you need to prove is true). But your factorization is correct. You should have tried factorizing the expression directly.Log in to reply

– Ayush Rai · 9 months, 4 weeks ago

Actually i did direct facotorization but make you'll understand some tricks,i wrote that way.Log in to reply

– Aditya Thomas · 9 months, 1 week ago

Awesome!!!!! I have a proof but it starts with assume the sides are a . a+d and a- dLog in to reply

– Ayush Rai · 9 months, 1 week ago

Thats actually not valid.Since it is only a special case of it.Log in to reply

– Neel Khare · 9 months, 1 week ago

how was NTSE ayushLog in to reply

Problem 6(b) is quite interesting. We can use Euler's theorem to prove the statement.

Let us first consider the case where \(\gcd(n,2)=\gcd(n,5)=1\), so \(\gcd(n,10)=1\). Now, from Euler's Theorem, we have \(10^{\phi(n)}\equiv 1\pmod{n}\).

We then consider the number \(f(n)=\sum\limits_{k=0}^{n-1}5\times 10^{k\phi(n)}\) which is just the number with \(5\) at the unit's digit, \((\phi(n)+1)^{\textrm{th}}\) digit from the right and so on till \(((n-1)\phi(n)+1)^{\textrm{th}}\) digit from the right and \(0\) at all the other digit places. Now, using Euler's theorem, we have,

\[f(n)\equiv\sum\limits_{k=0}^{n-1}5\times 1\equiv 5n\equiv 0\pmod{n}\]

So, we have our required number as \(f(n)\).

Now, for the general case where \(n\) isn't necessarily coprime to either \(2\) or \(5\), if \(n\) has \(2^{a_1}5^{a_2}\) in its unique prime factorization, then our required number is just \(f(n/(2^{a_1}5^{a_2}))\) appended with \((a_1+a_2)\) zeros (or you can also use \(\max(a_1,a_2)\) zeros). – Prasun Biswas · 9 months, 1 week ago

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6a is quite easy – Aditya Thomas · 9 months, 1 week ago

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– Ayush Rai · 9 months, 1 week ago

Only PPT required.Did u get second one?Log in to reply

– Prasun Biswas · 9 months, 1 week ago

I posted a solution for it. You can check it out in my comment below.Log in to reply

Did anyone get P5 ? – Alan Joel · 9 months, 3 weeks ago

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Problem 1a

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Problem \(4(b):\) I think the answer is only \((4,1).\)I dont have the proof.But my opinion is that there is only one possibility. – Ayush Rai · 9 months, 4 weeks ago

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Solution for \(3(b):\)

Let n be of the form \((2k+1)\) since they are all odd numbers.

\(n^2-n={(2k+1)}^2-(2k+1)=(2k+1)(2k)=2(2k+1)(k).\)Since every \(n^2-n\) has a common factor,therefore the greatest common divisor is \(\boxed{2}.\) – Ayush Rai · 9 months, 4 weeks ago

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\(6 (a)\) is simply pigeonhole principle based on modulus; at least 2 will leave same remainder when divided by \(n\).

Hence; difference between them will bea multiple of \(n\) – Yatin Khanna · 9 months, 4 weeks ago

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– Yatin Khanna · 9 months, 4 weeks ago

None. I did not give the test. I am just trying some questions. I got to know about the name of this test just now.Log in to reply

The answer to 1(a) is \(130<x <210\)

Correct? – Yatin Khanna · 9 months, 4 weeks ago

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– Alan Joel · 9 months, 4 weeks ago

I got x < 210 but I was unable to find the minimum valueLog in to reply

I can explain just with words..not much with equations and steps. – Yatin Khanna · 9 months, 4 weeks ago

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– Ayush Rai · 9 months, 4 weeks ago

Well the answer should be 190<x<210 according to me.Log in to reply

– Yatin Khanna · 9 months, 4 weeks ago

Oh wait I did not see that minus sign over there in second given equation; mistook for a plus sign.Log in to reply

– Svatejas Shivakumar · 9 months, 4 weeks ago

Actually, I just noticed that I made a typo(the plus sign) so you probably saw the old equation. Sorry for the mistake.Log in to reply

Junior includes what age range ? – Harsh Shrivastava · 9 months, 4 weeks ago

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– Svatejas Shivakumar · 9 months, 4 weeks ago

Class 9 and 10Log in to reply

– Kushagra Sahni · 9 months, 3 weeks ago

Any friend of yours who gave the Inter Paper? I want to see the inter one. I was selected for the Final but couldn't come so try to find out the paperLog in to reply

Is the answer for 4A 65% ? Also, in Q2, in triangle AEF, AD acts as altitude and angle bisector, implying AEF is isosceles, hence AE = AF. As F lies on perimeter of ABC, AF < AB. Using the fact that AB = AC and AE = AF, we get AE < AC contradicting the fact that AE lies on AC extended. Therefore, F coincides with B and E coincides with C. And now, I assumed BC to be 2x and I got MN = x^2/2016. Is my solution correct ? – Alan Joel · 9 months, 4 weeks ago

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– Yatin Khanna · 9 months, 3 weeks ago

In question 2 what is meant by the perpendicular to AD through D; doesnt that simply mean BC?Log in to reply

– Alan Joel · 9 months, 3 weeks ago

Yeah it is. The question seems to be wrong. Read my explanation on the top to know more.Log in to reply

– Ayush Rai · 9 months, 4 weeks ago

yes this equation is correct.even i got till here itself but the answer is numerical.Log in to reply