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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Answer to question 6.
Let $a=m+\frac{b}{c}$ where $m$ is any integer and $0<b<c$ .
Then $a(a-3{a})=( m+\frac{b}{c})( m-2\frac{b}{c})$$m^{2}-\frac{bm}{c}+\frac{2b^{2}}{c^{2}}$.
$m^{2}-\frac{2b^{2}-bcm}{c^{2}}$
Now, $m$ is an integer . Let's consider $\frac{2b^{2}-bcm}{c^{2}}=k$ where $k$ is an integer .
After solving we get $\frac{b}{c}=\frac{m+_{-}\sqrt{m^{2}+8k}}{4}$.....(I)
But $\frac{b}{c}<1$....(II)
Now putting value of $\frac{b}{c}$ from (I) to (II).
We get $m+k<2$
Therefore there are infinitely many integers $m,k$ such that $m+k<2$.
Hence proved.

@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.

Answer to question number 2.
$P_{1}(m)-P_{2}(n)=a_{1}-a_{2}=\frac{2(b_{2}-b_{1})}{m-n}$
This implies that $a_{1}-a_{2}=even$.
$P_{1}(m)-P_{2}(n)=a_{1}+a_{2}=-2(m+n)$
This implies that $a_{1}+a_{2}=even$.
Hence proved.

Answer to question 3.
$\frac{6l-1}{4l-3}=\frac{7k-5}{5k-3}$
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 $1,\frac{43}{31},\frac{5}{3},\frac{37}{27},\frac{13}{9},\frac{19}{13},\frac{61}{43},\frac{55}{39}$

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## Comments

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TopNewestwhen will the results of gmo will be declared

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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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I was able to do 4.5 questions do i stand a chance to get selected...

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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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Answer to question 6. Let $a=m+\frac{b}{c}$ where $m$ is any integer and $0<b<c$ . Then $a(a-3{a})=( m+\frac{b}{c})( m-2\frac{b}{c})$ $m^{2}-\frac{bm}{c}+\frac{2b^{2}}{c^{2}}$. $m^{2}-\frac{2b^{2}-bcm}{c^{2}}$ Now, $m$ is an integer . Let's consider $\frac{2b^{2}-bcm}{c^{2}}=k$ where $k$ is an integer . After solving we get $\frac{b}{c}=\frac{m+_{-}\sqrt{m^{2}+8k}}{4}$.....(I) But $\frac{b}{c}<1$....(II) Now putting value of $\frac{b}{c}$ 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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you can try the 6 question posted by me they all are of gmo.

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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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@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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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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Answer to question number 2. $P_{1}(m)-P_{2}(n)=a_{1}-a_{2}=\frac{2(b_{2}-b_{1})}{m-n}$ This implies that $a_{1}-a_{2}=even$. $P_{1}(m)-P_{2}(n)=a_{1}+a_{2}=-2(m+n)$ This implies that $a_{1}+a_{2}=even$. Hence proved.

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Shivam dud you gave GMO or RMO?

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RMO

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Answer to question 3. $\frac{6l-1}{4l-3}=\frac{7k-5}{5k-3}$ 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 $1,\frac{43}{31},\frac{5}{3},\frac{37}{27},\frac{13}{9},\frac{19}{13},\frac{61}{43},\frac{55}{39}$

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1 will be rejected as it isn't a fraction.

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Can anybody please post a proper solution for question $4$ with explanation?Log in to reply

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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