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).
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Overrated !
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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: 6 0 0 x + 7 0 0 y = 4 2 0 0 7 x + 6 y . G c d ( x , 6 0 0 ) = G c d ( y , 7 0 0 ) = 1 . Now,let us say that ( 7 x + 6 y ) ∗ a = 4 2 0 0 .Substituting this value in the fraction we got,we get a 1 .Now,we have to minimize a ,thus,we would have to maximize ( 7 x + 6 y ) .Now,first thing we get is that ( 7 x + 6 y ) can't be divisible by 7 or 6 .Now,using this fact keep ruling out the factors of 4 2 0 0 ,and in the end you will get that the max. value of ( 7 x + 6 y ) is 25.Hence, a is equal to 168.
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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 !
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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?
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@Adarsh Kumar – Have you understood Andrei's solution upto step 4 ? The factorisation of 6 × 7 0 0 = 6 × 7 × 4 × 2 5 and GCD( 6 × 7 × 4 , 7 q 1 + 6 q 2 ) = 1 made me take 2 5 ∣ 7 q 1 + 6 q 2 as a suitable constraint. We can see that minimum integral values of q 1 and q 2 such that 2 5 ∣ 7 q 1 + 6 q 2 is attained at least integral solution ( q 1 , q 2 ) for the diophantine equation 7 q 1 + 6 q 2 = 2 5 , and I think you very well know how to solve this ( A standard diophantine equation of form a x + b y = c ).We come to know from Euclid's algorithm that the least solution set obeying the equation is ( 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 ?
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@Venkata Karthik Bandaru – Yes,but I did not understand what encouraged him to do step 4.I mean why 6x7x4?
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@Adarsh Kumar – The encouragement is just the factorisation itself, nothing much !
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@Venkata Karthik Bandaru – Ok thanx buddy!
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@Adarsh Kumar – You are welcome :) !
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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...
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@A Former Brilliant Member – Sure bro, You too are awesome ! Add me on G+ !
@Adarsh Kumar – You are of class 9 ? Even my exams are over !
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@Venkata Karthik Bandaru – Yes I am.Are you on fb?
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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!
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@Venkata Karthik Bandaru – Can I add you in my circle?
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@Adarsh Kumar – Yeah sure :) !
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@Venkata Karthik Bandaru – Okay,so,same name on that too?
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@Adarsh Kumar – Venkata Karthik Bandaru is my name on G+. https://google.com+VenkataKarthikBandaru
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@Venkata Karthik Bandaru – Will start chatting!
@Venkata Karthik Bandaru – Got you! Thanx!
@Venkata Karthik Bandaru – Hey look,light just went out.Will contact u as soon as it comes back!
@Venkata Karthik Bandaru – You wanna chat on google+?
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@Adarsh Kumar – Yeah sure bro.
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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.
@Venkata Karthik Bandaru – Yes,now I'am clear.Thank you for explaining.
(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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This solution is not mine: g c d ( q 1 , 6 0 0 ) = 1 , g c d ( q 2 , 7 0 0 ) = 1
6 0 0 q 1 + 7 0 0 q 2 = 6 × 7 0 0 7 q 1 + 6 q 2
It's easy to see that g c d ( 6 , q 1 ) = 1 , g c d ( 7 , q 2 ) = 1
So g c d ( 6 × 7 × 4 , 7 q 1 + 6 q 2 ) = 1
To minimize the denominator , we try to find q 1 , q 2 such that 2 5 ∣ 7 q 1 + 6 q 2
( q 1 , q 2 ) = ( 1 , 3 ) is a choice.
6 0 0 1 + 7 0 0 3 = 1 6 8 1
The minimum value of the denominator is therefore 168.