Euler's Theorem: Euler's Totient Function

1 24 , 2 24 , 3 24 , , 23 24 , 24 24 \dfrac1{24} , \dfrac{2}{24}, \dfrac{3}{24}, \ldots , \dfrac{23}{24}, \dfrac{24}{24}

How many of the fractions above cannot be reduced?

6 8 10 12

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Brilliant Mathematics Staff
Aug 1, 2020

The fractions that cannot be reduced are those in which the numerator has no prime factors in common with 24, or equivalently, those in which the numerator is relatively prime with 24. Thus, the number of irreducible fractions is ϕ ( 24 ) \phi(24) , where ϕ \phi is Euler's totient function . Since the prime factors of 24 = 2 3 × 3 24=2^3\times 3 are 2 and 3 only, every integer that is a multiple of neither 2 nor 3 is relatively prime with 24.

Therefore, ϕ ( 24 ) = 24 ( 1 1 2 ) ( 1 1 3 ) = 8 , \phi(24) =24 \left( 1 - \dfrac12\right) \left( 1 - \dfrac13\right) = 8, and thus 8 of the 24 fractions are irreducible.

I never understood Euler's totient function. You just helped me by explaining it to me in 3 3 minutes! @Brilliant Mathematics

Yajat Shamji - 10 months, 2 weeks ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...