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Difference between revisions of "2013 Mock AIME I Problems/Problem 7"

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== Solution ==
 
== Solution ==
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<math>\boxed{005}</math>.
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Note that the only non-primitive <math>7</math>th or <math>9</math>th root of unity with a positive imaginary part is <math>e^{i\tfrac{6\pi}9} = e^{i\tfrac{2\pi}3}</math>. Listing the other such roots shows that both <math>S</math> and <math>T</math> have <math>3</math> elements, so <math>C</math> is equal to <math>3</math> times the sum of all the elements of <math>S</math> plus <math>3</math> times the sum of all the elements of <math>T</math>, because all of the terms are added thrice to the sum.
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Because all of the <math>7</math>th roots of unity sum to <math>0</math>, the sum of their real parts must be <math>0</math>. Without <math>1</math>, the sum of their real parts is <math>-1</math>. Because reciprocals of <math>7</math>th roots of unity are also <math>7</math>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 <math>-\tfrac1 2</math>. The same ''would'' be true for the <math>9</math>th roots of unity, but we have to remember to exclude <math>e^{i\tfrac{2\pi}3}</math>, which, by [[Euler's Identity]], has a real part of <math>\cos(\tfrac{2\pi}3)=-\tfrac 1 2</math>. Subtracting this from <math>-\tfrac 1 2</math> yields <math>0</math>, so the sum of the real parts of the elements of <math>T</math> is <math>0</math>.
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Thus, <math>|\Re(C)| = |3(-\tfrac 1 2)| = \tfrac 3 2</math>, so our answer is <math>3+2=\boxed{005}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 12:55, 30 July 2024

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

See also

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