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 of satisfy the equation :
Prove that is the arithmetic mean of .
2.In an isosceles , . The bisector of meets the side at . The line perpendicular to through meets at and produced at . Perpendiculars from and to are and respectively. If units, find the length of .
3.(a). Two circles with centres and and radii and respectively intersect each other at and and units. Chord is drawn to the bigger circle to cut it at and the smaller circle at such that is the midpoint of . Find the length of .
(b). Find the greatest common divisor of the numbers where
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 problems in Algebra, in Geometry and in Number Theory. He further found that he solved correctly of problems in Algebra and Number Theory put together, 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 such that .
5. are positive real numbers. Find the minimum value of
6.(a). Show that among any whole numbers, one can find two numbers such that their difference is divisible by .
(b). Show that for any natural number , there is a positive integer all of whose digits are or and is divisible by .
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Problem 1b
Log in to reply
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
Actually i did direct facotorization but make you'll understand some tricks,i wrote that way.
Awesome!!!!! I have a proof but it starts with assume the sides are a . a+d and a- d
Log in to reply
Thats actually not valid.Since it is only a special case of it.
Log in to reply
Junior includes what age range ?
Log in to reply
Class 9 and 10
Log in to reply
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
14 - 15 years
Problem 1a
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 10ϕ(n)≡1(modn).
We then consider the number f(n)=k=0∑n−15×10kϕ(n) which is just the number with 5 at the unit's digit, (ϕ(n)+1)th digit from the right and so on till ((n−1)ϕ(n)+1)th digit from the right and 0 at all the other digit places. Now, using Euler's theorem, we have,
f(n)≡k=0∑n−15×1≡5n≡0(modn)
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 2a15a2 in its unique prime factorization, then our required number is just f(n/(2a15a2)) appended with (a1+a2) zeros (or you can also use max(a1,a2) zeros).
The answer to 1(a) is 130<x<210
Correct?
Log in to reply
I got x < 210 but I was unable to find the minimum value
Log in to reply
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.
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
Solution for 3(b):
Let n be of the form (2k+1) since they are all odd numbers.
n2−n=(2k+1)2−(2k+1)=(2k+1)(2k)=2(2k+1)(k).Since every n2−n has a common factor,therefore the greatest common divisor is 2.
Did anyone get P5 ?
6a is quite easy
Log in to reply
Only PPT required.Did u get second one?
Log in to reply
I posted a solution for it. You can check it out in my comment below.
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 ?
Log in to reply
yes this equation is correct.even i got till here itself but the answer is numerical.
In question 2 what is meant by the perpendicular to AD through D; doesnt that simply mean BC?
Log in to reply
Yeah it is. The question seems to be wrong. Read my explanation on the top to know more.
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.
Log in to reply
https://math.stackexchange.com/questions/2065448/find-integral-solution-of-ab-ba-3