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almogsr Your solution is perfect.

There is a short-cut to find the co-primes of 100 that are less than 50

Totient(x)= \(x*(1-\frac{1}{a})(1-\frac{1}{b})\)....., where a,b are prime factors of x
totient(50)=\(50*(1-\frac{1}{2})(1-\frac{1}{5})\)=\(50*\frac{1}{2}*\frac{4}{5}\)=20


almogsr
since 0<a<1 we learn that x<y ==> combined with x+y=100 we get x=1,...,49 and y=99,...,50
there are 50 fractions we can create.
there are 25 pairs of (x,y) which are both even, therefore not at their lowest term form
there are 5 more pairs where x=10*k+5, such that we count all multiples of 5 but not the multiple of 10.
multiples of 3 are out of sync, so does the rest of the multiples of primes.
so we get 50-25-5=20.

please let me know if you agree with my solution.

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