Group Mathematical Olympiad is conducted by CBSE . It is only for CBSE students . Every CBSE school can send 5 students for it. Around 30 students are selected from the country and are eligible to write INMO.
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Aditya Chauhan
3 years, 8 months ago
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[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
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Top NewestAnswer to question 3. 4l−36l−1=5k−37k−5 After cross multiplication we get 8k+l+lk=6 Adding 8 to both sides 8k+8+l+lk=14 8(k+1)+l(1+k)=14 (8+l)(1+k)=14 Since l,k are integers therefore the solutions of (l,k)=(−1,1),(−6,6),(6,0),(−7,13),(−15,−3),(−22,−2),(−10,−8),(−9,−15) Therefore the required fractions are 1,3143,35,2737,913,1319,4361,3955
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Can anybody please post a proper solution for question 4 with explanation?
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Number of ways of selecting points from points is 28C3=3276
Now, we will eliminate some conditions..
Number of ways of selecting all 3 as adjacent points is 28.
Number of ways of selecting 2 adjacent points and one not adjacent with them is 28×24=672
Number of ways of selecting two points opposite diametrically along with the third point not adjacent to the former points is 14×22=308.
Hence, the desired result will be 3276-28-672-308=2268
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1 will be rejected as it isn't a fraction.
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Answer to question number 2. P1(m)−P2(n)=a1−a2=m−n2(b2−b1) This implies that a1−a2=even. P1(m)−P2(n)=a1+a2=−2(m+n) This implies that a1+a2=even. Hence proved.
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Shivam dud you gave GMO or RMO?
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RMO
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How many did you solve?
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5
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when will the results of gmo will be declared
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Can anyone tell the answer of 5th question
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it is 4/3
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I m getting 1/3
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see i can't post my solution because it is too long but you can verify it by construction.
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I was able to do 4.5 questions do i stand a chance to get selected...
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Is the answer to question 4 2268?
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Yes it is
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Can you tell me what is the expected cutoff for gmo?
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Answer to question 6. Let a=m+cb where m is any integer and 0<b<c . Then a(a−3a)=(m+cb)(m−2cb) m2−cbm+c22b2. m2−c22b2−bcm Now, m is an integer . Let's consider c22b2−bcm=k where k is an integer . After solving we get cb=4m+−m2+8k.....(I) But cb<1....(II) Now putting value of cb from (I) to (II). We get m+k<2 Therefore there are infinitely many integers m,k such that m+k<2. Hence proved.
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Bro ur equation which is quadratic in m is wrong also we will get 4 values of a
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A general solution a=(2n+1)+0.5 where n is a integer.
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@Shivam Jadhav This ques is a little different from the RMO one , we have to find all values of 'a' between 3 & 4. @Devansh Shah is right , there will be 4 values.
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m cant be 'any' integer as 3<a<4. So a will be 3.something or a = 3 + b/c
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you can try the 6 question posted by me they all are of gmo.
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