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 $9$ 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 $3+2=\boxed{005}$ .

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2013 Mock AIME I Problems/Problem 7

Problem

Solution

See also