There are two positive integers a and b such that:
a 1 + b 1 = 2 0 1
Now, what is the highest value of a + b ?
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@Shrenik Jobanputra Your solution doesn't explain why it's the highest value...
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The thing is that I just did the competition just 2 days ago. I know the answer but I still don't know the procedure of working it out. If possible @Satvik Golechha , can U PLEASE also write a solution.
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@Shrenik Jobanputra my solution was listing out all the possible pairs and crossing out. Not bash-free though.
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@Satvik Golechha – That is actually what I would also do.
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@Shrenik Jobanputra – Well I failed that pretty badly. Tried to find denominators from 20 to 40 that worked out. I found as many as I could and picked the pair with the highest sum. Unfortunately that didn't really work. Looking back on it, I think I over complicated it a bit, although I'm not sure what a simpler way would be.
Lets start with 20x2/1=40
1/40 + 1/40 = 1/20
Now 20x3/2=30
1/30 + 1/60 = 1/20
20x4/3 is not an integer
20x5/4=25
1/25 + 1/100 = 1/20
& so on till 20x21/20=21
So 1/21 + 1/420 = 1/20
The trick here is that closer we are to 1/20, smaller the value(& hence greater the denominator) we will have to add to reach 1/20
21 is the closest to 20 & hence the result.
There is an easier way to reach that final stage. Use the identity: 1/x = 1/(x+1) + 1/(x)(x+1). This can be proven by either using the Principle of Mathematical Induction or expanding the RHS. This means that 1/20 = 1/21 + 1/(20)(21), i.e. 1/20 = 1/21 + 1/420. Like you said in your solution, the closer you are to 20, the smaller the value and hence the greater the denominator which we will need to add to obtain 1/20/ So a+b = 21+ 420 = 441
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