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.

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 \[ 1,\frac{43}{31},\frac{5}{3},\frac{37}{27},\frac{13}{9},\frac{19}{13},\frac{61}{43},\frac{55}{39}\]

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 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.

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) 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

Sort by:

TopNewestAnswer 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 \[ 1,\frac{43}{31},\frac{5}{3},\frac{37}{27},\frac{13}{9},\frac{19}{13},\frac{61}{43},\frac{55}{39}\]

Log in to reply

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

Log in to reply

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

Log in to reply

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.

Log in to reply

Shivam dud you gave GMO or RMO?

Log in to reply

RMO

Log in to reply

Log in to reply

Log in to reply

when will the results of gmo will be declared

Log in to reply

Is the answer to question 4 2268?

Log in to reply

Yes it is

Log in to reply

Can you tell me what is the expected cutoff for gmo?

Log in to reply

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

Log in to reply

Can anyone tell the answer of 5th question

Log in to reply

it is 4/3

Log in to reply

I m getting 1/3

Log in to reply

Log in to reply

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.

Log in to reply

@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.

Log in to reply

Bro ur equation which is quadratic in m is wrong also we will get 4 values of a

Log in to reply

A general solution \[a=(2n+1)+0.5\] where \(n\) is a integer.

Log in to reply

you can try the 6 question posted by me they all are of gmo.

Log in to reply

m cant be 'any' integer as 3<a<4. So a will be 3.something or a = 3 + b/c

Log in to reply