GMO 2015 problem

Find the number of fractions that can be written simultaneously in the forms 7 k 5 5 k 3 \frac{7k-5}{5k-3} and 6 l 1 4 l 3 \frac{6l-1}{4l-3} for some integers k , l k,l .


The answer is 8.

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1 solution

Shivam Jadhav
Dec 16, 2015

6 l 1 4 l 3 = 7 k 5 5 k 3 \frac{6l-1}{4l-3}=\frac{7k-5}{5k-3} After cross multiplication we get 8 k + l + l k = 6 8k+l+lk=6 Adding 8 to both sides 8 k + 8 + l + l k = 14 8k+8+l+lk=14 8 ( k + 1 ) + l ( 1 + k ) = 14 8(k+1)+l(1+k)=14 ( 8 + l ) ( 1 + k ) = 14 (8+l)(1+k)=14 Since l , k 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 ) (l,k)=(-1,1),(-6,6),(6,0),(-7,13),(-15,-3),(-22,-2),(-10,-8),(-9,-15) Therefore the required fractions are 1 , 43 31 , 5 3 , 37 27 , 13 9 , 19 13 , 61 43 , 55 39 1,\frac{43}{31},\frac{5}{3},\frac{37}{27},\frac{13}{9},\frac{19}{13},\frac{61}{43},\frac{55}{39}

I did the same thing in GMO!

Adarsh Kumar - 5 years, 6 months ago

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Me too. You gave GMO? How much do you expect? @Adarsh Kumar

Aditya Chauhan - 5 years, 6 months ago

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How can we appear for GMO?

Kushagra Sahni - 5 years, 5 months ago

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@Kushagra Sahni Every CBSE school can send upto 5 children for it. So if you are in a CBSE school, then you should consult your school teachers.

Aditya Chauhan - 5 years, 5 months ago

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@Aditya Chauhan Eligibility is 9th to 11th?

Kushagra Sahni - 5 years, 5 months ago

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@Kushagra Sahni Students upto class 11th can give it. Maybe 8th class is also eligible. But I am sure of 9-11.

Aditya Chauhan - 5 years, 5 months ago

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@Aditya Chauhan Which is easier RMO or GMO?

A Former Brilliant Member - 5 years, 5 months ago

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@A Former Brilliant Member Dude,both are the same thing,just name is different because GMO is RMO for CBSE schools.

Adarsh Kumar - 5 years, 5 months ago

Sorry for the late reply,haven't veen much active on brilliant for sometimes.I got 4 perfect and the rest nil.

Adarsh Kumar - 5 years, 5 months ago

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