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NMTC Junior Level Final Test 2016

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 $$n$$.

Note by Svatejas Shivakumar
9 months, 4 weeks ago

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Problem 1b

· 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. · 9 months, 4 weeks ago

Actually i did direct facotorization but make you'll understand some tricks,i wrote that way. · 9 months, 4 weeks ago

Awesome!!!!! I have a proof but it starts with assume the sides are a . a+d and a- d · 9 months, 1 week ago

Thats actually not valid.Since it is only a special case of it. · 9 months, 1 week ago

how was NTSE ayush · 9 months, 1 week ago

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). · 9 months, 1 week ago

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

Only PPT required.Did u get second one? · 9 months, 1 week ago

I posted a solution for it. You can check it out in my comment below. · 9 months, 1 week ago

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

Problem 1a

· 9 months, 3 weeks ago

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. · 9 months, 4 weeks ago

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 $$\boxed{2}.$$ · 9 months, 4 weeks ago

$$6 (a)$$ 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 $$n$$ · 9 months, 4 weeks ago

Comment deleted 9 months 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. · 9 months, 4 weeks ago

The answer to 1(a) is $$130<x <210$$
Correct? · 9 months, 4 weeks ago

I got x < 210 but I was unable to find the minimum value · 9 months, 4 weeks ago

If you take $$b$$ arbitrarily close to 0; the expression will come closer and closer towards 130; but since $$b$$ is positive; clearly 130 is strictly less than the given expression.
I can explain just with words..not much with equations and steps. · 9 months, 4 weeks ago

Comment deleted 9 months ago

Well the answer should be 190<x<210 according to me. · 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. · 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. · 9 months, 4 weeks ago

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

Class 9 and 10 · 9 months, 4 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 paper · 9 months, 3 weeks ago

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 ? · 9 months, 4 weeks ago

In question 2 what is meant by the perpendicular to AD through D; doesnt that simply mean BC? · 9 months, 3 weeks ago

Yeah it is. The question seems to be wrong. Read my explanation on the top to know more. · 9 months, 3 weeks ago