Brainsteem Quickfire questions is a series of slightly easier, certainly faster, mathematical challenges in the Brainsteem family!
Winner will get 2 SBD or a @randowhale upvote for free!
Second prize is 1 SBD for free selected randomly from all other correct solutions!
Let's write the fraction 1/20 as the sum of two other unit fractions:
1/20 = 1/a + 1/b , such that a and b are positive integers and a < b.
Find the smallest possible sum (a+b).
Easy!
Prizes
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The First Prize will be 2 SBD or the chance of a @randowhale upvote.
The Second Prize will be 1 SBD to any other correct answer in the comments section, chosen at random.
Prizes will be awarded after the 7 days have expired. Winners will be announced at the next puzzle after the draw.
The main thing is to enjoy the challenge and maybe learn something new.
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This is again the sort of question where the solver needs a method to reduce the possible space of answers.
The smallest possible sum (a+b)= 36+45=81
1/20= 1/a+1/b= (a+b)/ab --> 20(a+b)=ab
b=20a/(a-20)=20+400/(a-20)
Hence a+b=a+20-400/(a-20),
By differentiation, a+b is minimized when (a-20)^2=400, i.e. a=40, b=40.
However, this does not fit the requirement a<b. The solution pair that is closest to this pair would be the solution.
Consider b=20+400/(a-20) and b is an integer,
(a-20) has to be a factor of 400. As 16 is a factor of 400 and none of 17,18,19 is factor of 400, we can see that (a,b)=(36,45) is the pair of solution we are looking for.
That's too hard for me :p
a=33
b=49
You have to include a valid method in your comment
Yes, what is your method?
Sometimes, writing out the method will reveal an error.
I know that writing maths on a computer can be a pain - some people just do it on paper and post a photo of it - that's fine.