All-Russian Olympiad Problem 5

The denominators of two irreducible fractions are 600 and 700. Find the minimum value of the denominator of their sum (written as an irreducible fraction).

This problem is part of this set .


The answer is 168.

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2 solutions

This solution is not mine: g c d ( q 1 , 600 ) = 1 , g c d ( q 2 , 700 ) = 1 gcd(q_1,600)=1, gcd(q_2,700)=1

q 1 600 + q 2 700 = 7 q 1 + 6 q 2 6 × 700 \frac{q_1}{600}+\frac{q_2}{700}=\frac{7q_1+6q_2}{6 \times 700}

It's easy to see that g c d ( 6 , q 1 ) = 1 , g c d ( 7 , q 2 ) = 1 gcd(6, q_1)=1, gcd(7, q_2)=1

So g c d ( 6 × 7 × 4 , 7 q 1 + 6 q 2 ) = 1 gcd (6 \times 7 \times 4, 7q_1+6q_2)=1

To minimize the denominator , we try to find q 1 , q 2 q_1,q_2 such that 25 7 q 1 + 6 q 2 25|7q_1+6q_2

( q 1 , q 2 ) = ( 1 , 3 ) (q_1,q_2)=(1,3) is a choice.

1 600 + 3 700 = 1 168 \frac{1}{600}+\frac{3}{700}=\frac{1}{168}

The minimum value of the denominator is therefore 168.

Overrated !

Venkata Karthik Bandaru - 6 years, 2 months ago

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How did you solve it? Though I didn't solve it(I just saw the solution),I came up with a rather easy method: x 600 + y 700 = 7 x + 6 y 4200 . G c d ( x , 600 ) = G c d ( y , 700 ) = 1. \dfrac{x}{600}+\dfrac{y}{700}=\dfrac{7x+6y}{4200}.Gcd(x,600)=Gcd(y,700)=1. Now,let us say that ( 7 x + 6 y ) a = 4200 (7x+6y)*a=4200 .Substituting this value in the fraction we got,we get 1 a \dfrac{1}{a} .Now,we have to minimize a a ,thus,we would have to maximize ( 7 x + 6 y ) (7x+6y) .Now,first thing we get is that ( 7 x + 6 y ) (7x+6y) can't be divisible by 7 7 or 6 6 .Now,using this fact keep ruling out the factors of 4200 4200 ,and in the end you will get that the max. value of ( 7 x + 6 y ) (7x+6y) is 25.Hence, a a is equal to 168.

Adarsh Kumar - 6 years, 2 months ago

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I just solved the way Andrei did, it was the first one to come up to my mind. A very straight forward one. Your approach is nice too !

Venkata Karthik Bandaru - 6 years, 2 months ago

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@Venkata Karthik Bandaru Yeah thanx!But what encouraged you to take 6x7x4,and how can you prove that it is the minimum? @Karthik Venkata could you please explain Andrei's solution to me?

Adarsh Kumar - 6 years, 2 months ago

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@Adarsh Kumar Have you understood Andrei's solution upto step 4 ? The factorisation of 6 × 700 = 6 × 7 × 4 × 25 6 \times 700 = 6 \times 7 \times 4 \times 25 and GCD( 6 × 7 × 4 , 7 q 1 + 6 q 2 6 \times 7 \times 4, 7q_{1}+6q_{2} ) = 1 made me take 25 7 q 1 + 6 q 2 25|7q_{1}+6q_{2} as a suitable constraint. We can see that minimum integral values of q 1 q_{1} and q 2 q_{2} such that 25 7 q 1 + 6 q 2 25|7q_{1}+6q_{2} is attained at least integral solution ( q 1 , q 2 q_{1}, q_{2} ) for the diophantine equation 7 q 1 + 6 q 2 = 25 7q_{1}+6q_{2}=25 , and I think you very well know how to solve this ( A standard diophantine equation of form a x + b y = c ax+by=c ).We come to know from Euclid's algorithm that the least solution set obeying the equation is ( q 1 , q 2 ) (q_{1}, q_{2}) = ( 1,3 ). [ If you do not know how to solve diophantine equations, look for it in Number theory Wikis in Brilliant]. Are you clear now ?

Venkata Karthik Bandaru - 6 years, 2 months ago

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@Venkata Karthik Bandaru Yes,but I did not understand what encouraged him to do step 4.I mean why 6x7x4?

Adarsh Kumar - 6 years, 2 months ago

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@Adarsh Kumar The encouragement is just the factorisation itself, nothing much !

Venkata Karthik Bandaru - 6 years, 2 months ago

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@Venkata Karthik Bandaru Ok thanx buddy!

Adarsh Kumar - 6 years, 2 months ago

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@Adarsh Kumar You are welcome :) !

Venkata Karthik Bandaru - 6 years, 2 months ago

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@Venkata Karthik Bandaru @Karthik Venkata could I chat as well? I am Andrei Golovanov with an IMO logo profile picture. I added you to my Math discusions circle! Do you allow me to chat? You just sound cool man, aiming for IMO and all...

A Former Brilliant Member - 6 years, 2 months ago

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@A Former Brilliant Member Sure bro, You too are awesome ! Add me on G+ !

Venkata Karthik Bandaru - 6 years, 2 months ago

@Adarsh Kumar You are of class 9 ? Even my exams are over !

Venkata Karthik Bandaru - 6 years, 2 months ago

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@Venkata Karthik Bandaru Yes I am.Are you on fb?

Adarsh Kumar - 6 years, 2 months ago

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@Adarsh Kumar Nope, but I am on Google+. I wonder how come you are such a brilliant guy ! Dont you feel confused about Combinatorics ? COMBNATORICS MAKES ME DIZZY ! You are a level 5 on Comb. Awesome!

Venkata Karthik Bandaru - 6 years, 2 months ago

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@Venkata Karthik Bandaru Can I add you in my circle?

Adarsh Kumar - 6 years, 2 months ago

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@Adarsh Kumar Yeah sure :) !

Venkata Karthik Bandaru - 6 years, 2 months ago

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@Venkata Karthik Bandaru Okay,so,same name on that too?

Adarsh Kumar - 6 years, 2 months ago

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@Adarsh Kumar Venkata Karthik Bandaru is my name on G+. https://google.com+VenkataKarthikBandaru

Venkata Karthik Bandaru - 6 years, 2 months ago

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@Venkata Karthik Bandaru Will start chatting!

Adarsh Kumar - 6 years, 2 months ago

@Venkata Karthik Bandaru Got you! Thanx!

Adarsh Kumar - 6 years, 2 months ago

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@Adarsh Kumar Got you too buddy !

Venkata Karthik Bandaru - 6 years, 2 months ago

@Venkata Karthik Bandaru Hey look,light just went out.Will contact u as soon as it comes back!

Adarsh Kumar - 6 years, 2 months ago

@Venkata Karthik Bandaru You wanna chat on google+?

Adarsh Kumar - 6 years, 2 months ago

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@Adarsh Kumar Yeah sure bro.

Venkata Karthik Bandaru - 6 years, 2 months ago

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@Venkata Karthik Bandaru Okay,I am unable to start chat.Can you start it,the name is the same.I have a really slow internet connection.

Adarsh Kumar - 6 years, 2 months ago

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@Adarsh Kumar Ok ,wait a min.

Venkata Karthik Bandaru - 6 years, 2 months ago

@Venkata Karthik Bandaru Yes,now I'am clear.Thank you for explaining.

Adarsh Kumar - 6 years, 2 months ago
Siddharth Iyer
Mar 22, 2015

(a/600)+(b/700)=(7a+6b)/((7)(2^3)(3)(5^2)). a is not a multiple of 2,3 and 5. b is not a multiple of 2,5 and 7. We see that 7a + 6b cannot be a multiple of 7 or 3 or 2. It could be a multiple of 25. Now by euclidean algorithm we know that it is possible to solve 7a+6b=25 as 25/gcd(7,6) is an integer. hence the result is 7*(2^3)(3) = 168

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