RMO Practice Question 2

Only the answer

$|(-1)+(-1)+0+(-1)+(-1)+1+(-1)+(-1)+1+(-1)+0+0+(-1)+(-1)+\\+1+1+(-1)+1+(-1)+(-1)+0+(-1)+0+1+(-1)+1|=|-7|=7$

Complete solution

• $\mu(n)=1$ if $n$ is a square-free positive integer with an even number of prime factors.
• $\mu(n)=-1$ if $n$ is a square-free positive integer with an odd number of prime factors.
• $\mu(n)=0$ if $n$ has a squared prime factor.

We should see the sum, so we should find only the square free numbers.

The square free numbers are(until $(2\times26-1=51)$ ):

$\texttt{1,2,3,5,6,7,10,11,13,14,15,17,19,21,22,23,26,29,30,31,33,34,35,37,38,39,41,42,43,46,47,51}$

The $2n-1$ sequence is the list of odd numbers. Without even numbers:

$\texttt{1,3,5,7,11,13,15,17,19,21,23,29,31,33,35,37,39,41,43,47,51 - 20 numbers}$

The numbers with odd prime factors are the prime numbers(because the smallest number with 3 prime factors is $3\times5\times7>51$ ). Prime numbers

$\texttt{3,5,7,11,13,17,19,23,29,31,37,41,43,47 - 13 numbers}$

The difference is $20-13=7$ numbers, so the solution is $7$ .