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- Project Euler #157: Solving the diophantine equation 1/a +1/b = p/10^n
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# Project Euler #157: Solving the diophantine equation 1/a +1/b = p/10^n

# Project Euler #157: Solving the diophantine equation 1/a +1/b = p/10^n

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I don't see how there could be multiple solutions (a,b,p) for different alpha's. 1/a+1/b is unique and p/(p1^alpha1*p2^alpha2) is also unique.

Why is the editorial available only after solving the problem? If anyone solves the problem why he would even look at the editorial. And even testcases aren't availavle :-(

hi there, I just want to ask is p must b prime number?

Hi,

great task. Thanks. I have two questions:

1.) you write, the "tuple" {a, b, p} has to be counted, but mathematically you write it as a set. How do I have to count then? Like it's a set, or like its a tuple?

For example if I have a solution:

a=2, b=5, p=7

then of course also

a=5, b=2, p=7 would be a solution, but the latter is not countet, because of the constraint a<=b, but what if I also have a solution

a=2, a=7, p=5

with the same alpha-values? Would I have to count these three as 1 solution (like if it's a set, or as two, like it is a tuple)?

If you mean they have to be counted as two distinct solutions you could consider restating the question as:

Count the number of distinct tuples with (a, b, p, alpha1, alpha2) where a<=b.

That would maybe be clearer and shorter.

2.) and related, p has to be a prime, right?

Thank you in advance JÃ¼rgen