2013 Mock AIME I Problems/Problem 7

Problem

Let $S$ be the set of all $7$th primitive roots of unity with imaginary part greater than $0$. Let $T$ be the set of all $9$th primitive roots of unity with imaginary part greater than $0$. (A primitive $n$th root of unity is a $n$th root of unity that is not a $k$th root of unity for any $1 \le k < n$.)Let $C=\sum_{s\in S}\sum_{t\in T}(s+t)$. The absolute value of the real part of $C$ can be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime numbers. Find $m+n$.

Solution

Note that the only non-primitive $7$th or $9$th root of unity with a positive imaginary part is $e^{i\tfrac{6\pi}9} = e^{i\tfrac{2\pi}3}$. Listing the other such roots shows that both $S$ and $T$ have $3$ elements, so $C$ is equal to $3$ times the sum of all the elements of $S$ plus $3$ times the sum of all the elements of $T$, because all of the terms are added thrice to the sum.

Because all of the $7$th roots of unity sum to $0$, the sum of their real parts must be $0$. Without $1$, the sum of their real parts is $-1$. Because reciprocals of $7$th roots of unity are also $7$th roots of unity (but with opposite imaginary parts and the same real part), the sum of the real parts of the roots with a positive imaginary part must be $-\tfrac1 2$. The same would be true for the primitive $9$th roots of unity, but we have to remember to exclude $e^{i\tfrac{2\pi}3}$, which, by Euler's Identity, has a real part of $\cos(\tfrac{2\pi}3)=-\tfrac 1 2$. Subtracting this from $-\tfrac 1 2$ yields $0$, so the sum of the real parts of the elements of $T$ is $0$.

Thus, $|\Re(C)| = |3(-\tfrac 1 2)| = \tfrac 3 2$, so our answer is $3+2=\boxed{005}$.

See also