NMTC Junior Level Final Test 2016

1.(a). If a,b,ca,b,c are positive reals and a+b+c=50a+b+c=50 and 3a+bc=703a+b-c=70. If x=5a+4b+2cx=5a+4b+2c, find the range of the values of xx.

(b). The sides a,b,ca,b,c of ΔABC\Delta ABC satisfy the equation : a2+2b2+2016c23ab4033bc+2017ac=0a^2+2b^2+2016c^2-3ab-4033bc+2017ac=0

Prove that bb is the arithmetic mean of a,ca,c.

2.In an isosceles ΔABC\Delta ABC, AB=ACAB=AC. The bisector ADAD of A\angle A meets the side BCBC at DD. The line perpendicular to ADAD through DD meets ABAB at FF and ACAC produced at EE. Perpendiculars from BB and DD to ACAC are BMBM and DNDN respectively. If AE=2016AE=2016 units, find the length of MNMN.

3.(a). Two circles with centres PP and QQ and radii 2\sqrt2 and 11 respectively intersect each other at AA and DD and PQ=2PQ=2 units. Chord ACAC is drawn to the bigger circle to cut it at CC and the smaller circle at BB such that BB is the midpoint of ACAC. Find the length of ACAC.

(b). Find the greatest common divisor of the numbers n2nn^2-n where n=3,5,7,9,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%50 \% problems in Algebra, 70%70 \% in Geometry and 80%80 \% in Number Theory. He further found that he solved correctly 62%62 \% of problems in Algebra and Number Theory put together, 74%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)(a,b) such that abba=3a^b-b^a=3.

5.a,b,ca,b,c are positive real numbers. Find the minimum value of a+3ca+2b+c+4ba+b+2c8ca+b+3c\dfrac{a+3c}{a+2b+c}+\dfrac{4b}{a+b+2c}-\dfrac{8c}{a+b+3c}

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

(b). Show that for any natural number nn, there is a positive integer all of whose digits are 55 or 00 and is divisible by nn.

Note by A Former Brilliant Member
3 years ago

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Problem 1b Problem 1b

Ayush G Rai - 3 years ago

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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.

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

Ayush G Rai - 3 years ago

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

Aditya Thomas - 3 years ago

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Thats actually not valid.Since it is only a special case of it.

Ayush G Rai - 3 years ago

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@Ayush G Rai how was NTSE ayush

Neel Khare - 3 years ago

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Junior includes what age range ?

Harsh Shrivastava - 3 years ago

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Class 9 and 10

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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 paper

Kushagra Sahni - 3 years ago

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Problem 1a Problem 1a

Ayush G Rai - 3 years ago

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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\gcd(n,2)=\gcd(n,5)=1, so gcd(n,10)=1\gcd(n,10)=1. Now, from Euler's Theorem, we have 10ϕ(n)1(modn)10^{\phi(n)}\equiv 1\pmod{n}.

We then consider the number f(n)=k=0n15×10kϕ(n)f(n)=\sum\limits_{k=0}^{n-1}5\times 10^{k\phi(n)} which is just the number with 55 at the unit's digit, (ϕ(n)+1)th(\phi(n)+1)^{\textrm{th}} digit from the right and so on till ((n1)ϕ(n)+1)th((n-1)\phi(n)+1)^{\textrm{th}} digit from the right and 00 at all the other digit places. Now, using Euler's theorem, we have,

f(n)k=0n15×15n0(modn)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)f(n).

Now, for the general case where nn isn't necessarily coprime to either 22 or 55, if nn has 2a15a22^{a_1}5^{a_2} in its unique prime factorization, then our required number is just f(n/(2a15a2))f(n/(2^{a_1}5^{a_2})) appended with (a1+a2)(a_1+a_2) zeros (or you can also use max(a1,a2)\max(a_1,a_2) zeros).

Prasun Biswas - 3 years ago

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The answer to 1(a) is 130<x<210130<x <210
Correct?

Yatin Khanna - 3 years ago

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I got x < 210 but I was unable to find the minimum value

Alan Joel - 3 years ago

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If you take bb arbitrarily close to 0; the expression will come closer and closer towards 130; but since bb is positive; clearly 130 is strictly less than the given expression.
I can explain just with words..not much with equations and steps.

Yatin Khanna - 3 years ago

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

Hence; difference between them will bea multiple of nn

Yatin Khanna - 3 years ago

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Solution for 3(b):3(b):
Let n be of the form (2k+1)(2k+1) since they are all odd numbers.
n2n=(2k+1)2(2k+1)=(2k+1)(2k)=2(2k+1)(k).n^2-n={(2k+1)}^2-(2k+1)=(2k+1)(2k)=2(2k+1)(k).Since every n2nn^2-n has a common factor,therefore the greatest common divisor is 2.\boxed{2}.

Ayush G Rai - 3 years ago

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Did anyone get P5 ?

Alan Joel - 3 years ago

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6a is quite easy

Aditya Thomas - 3 years ago

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Only PPT required.Did u get second one?

Ayush G Rai - 3 years ago

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

Prasun Biswas - 3 years ago

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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 - 3 years ago

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

Ayush G Rai - 3 years ago

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

Yatin Khanna - 3 years ago

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

Alan Joel - 3 years ago

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Problem 4(b):4(b): I think the answer is only (4,1).(4,1).I dont have the proof.But my opinion is that there is only one possibility.

Ayush G Rai - 3 years ago

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https://math.stackexchange.com/questions/2065448/find-integral-solution-of-ab-ba-3

vivek v - 1 year, 4 months ago

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