Question

If $\omega$ is the imaginary cube root of unity such that $|\left(\sum_{r=1}^{n}\left(r \cdot \sum_{p=1}^{r} \omega^{p-1}\right)\right)-155\omega|=\left(\sum_{r=1}^{n}\left(r \cdot \sum_{p=1}^{r} \omega^{p-1}\right)\right)-155\omega$, then n is equal to -

• A. 29
• B. 30
• C. 31
• D. 32

Answer 1

Answer: (A), (B), (C)

29

Answer 2

Let the given condition be $|Z|=Z$, where $Z = \left(\sum_{r=1}^{n}\left(r \cdot \sum_{p=1}^{r} \omega^{p-1}\right)\right)-155\omega$. This condition implies that $Z$ must be a non-negative real number, i.e., $Im(Z)=0$ and $Re(Z) \ge 0$.

First, simplify the inner sum $S_r = \sum_{p=1}^{r} \omega^{p-1}$. Using the property $1+\omega+\omega^2=0$:

• If $r \equiv 1 \pmod 3$, $S_r = 1$.
• If $r \equiv 2 \pmod 3$, $S_r = 1+\omega = -\omega^2$.
• If $r \equiv 0 \pmod 3$, $S_r = 1+\omega+\omega^2 = 0$.

Let $X = \sum_{r=1}^{n}\left(r \cdot S_r\right)$. We need $Im(X - 155\omega) = 0$, which means $Im(X) = 155 Im(\omega)$. Since $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$, $Im(\omega) = \frac{\sqrt{3}}{2}$. So, $Im(X) = 155 \frac{\sqrt{3}}{2}$.

Let's analyze $X$ based on $n \pmod 3$. We can express $X$ in terms of $k$, where $k = \lfloor (n-1)/3 \rfloor$.

For $n=3k$: $X_{3k} = \sum_{j=0}^{k-1} ((3j+1)S_{3j+1} + (3j+2)S_{3j+2} + (3j+3)S_{3j+3})$ $X_{3k} = \sum_{j=0}^{k-1} ( (3j+1) \cdot 1 + (3j+2) \cdot (-\omega^2) + (3j+3) \cdot 0 )$ $X_{3k} = \sum_{j=0}^{k-1} (3j+1 - (3j+2)\omega^2) = \sum_{j=0}^{k-1} (3j(1-\omega^2) + (1-2\omega^2))$ Using $1-\omega^2 = 2+\omega$ and $1-2\omega^2 = 3+2\omega$: $X_{3k} = \sum_{j=0}^{k-1} (3j(2+\omega) + (3+2\omega)) = 3(2+\omega)\frac{k(k-1)}{2} + k(3+2\omega)$ $X_{3k} = \frac{k}{2} (6k + (3k+1)\omega)$. $Im(X_{3k}) = \frac{k(3k+1)\sqrt{3}}{4}$. Equating $Im(X_{3k}) = 155 \frac{\sqrt{3}}{2}$: $\frac{k(3k+1)\sqrt{3}}{4} = 155 \frac{\sqrt{3}}{2} \implies k(3k+1) = 310 \implies 3k^2+k-310=0$. Solving for $k$: $k = \frac{-1 \pm \sqrt{1-4(3)(-310)}}{6} = \frac{-1 \pm \sqrt{1+3720}}{6} = \frac{-1 \pm \sqrt{3721}}{6} = \frac{-1 \pm 61}{6}$. Since $k$ must be a positive integer, $k = \frac{-1+61}{6} = 10$. If $k=10$, then $n=3k=30$. For $n=30$, $X_{30} = \frac{10}{2} (60+31\omega) = 300+155\omega$. $Z = X_{30} - 155\omega = (300+155\omega) - 155\omega = 300$. Since $Z=300 \ge 0$, $n=30$ is a valid solution.

For $n=3k+1$: $X_{3k+1} = X_{3k} + (3k+1)S_{3k+1} = X_{3k} + (3k+1) \cdot 1$. $Im(X_{3k+1}) = Im(X_{3k}) + Im(3k+1) = Im(X_{3k})$. So, $k=10$ still holds. If $k=10$, then $n=3k+1 = 3(10)+1 = 31$. For $n=31$, $X_{31} = X_{30} + 31 = (300+155\omega) + 31 = 331+155\omega$. $Z = X_{31} - 155\omega = (331+155\omega) - 155\omega = 331$. Since $Z=331 \ge 0$, $n=31$ is a valid solution.

For $n=3k+2$: $X_{3k+2} = X_{3k+1} + (3k+2)S_{3k+2} = X_{3k+1} + (3k+2)(-\omega^2)$. $Im(X_{3k+2}) = Im(X_{3k+1}) + Im(-(3k+2)\omega^2) = Im(X_{3k}) + Im((3k+2)(1+\omega))$. $Im(X_{3k+2}) = \frac{k(3k+1)\sqrt{3}}{4} + (3k+2)\frac{\sqrt{3}}{2}$. Equating $Im(X_{3k+2}) = 155 \frac{\sqrt{3}}{2}$: $\frac{k(3k+1)}{2} + (3k+2) = 155 \implies k(3k+1) + 2(3k+2) = 310$ $3k^2+k+6k+4 = 310 \implies 3k^2+7k-306=0$. Solving for $k$: $k = \frac{-7 \pm \sqrt{7^2-4(3)(-306)}}{6} = \frac{-7 \pm \sqrt{49+3672}}{6} = \frac{-7 \pm \sqrt{3721}}{6} = \frac{-7 \pm 61}{6}$. Since $k$ must be a positive integer, $k = \frac{-7+61}{6} = 9$. If $k=9$, then $n=3k+2 = 3(9)+2 = 29$. For $n=29$, $X_{29} = X_{3(9)} + (3(9)+1) + (3(9)+2)(-\omega^2)$. $X_{27} = \frac{9}{2}(6(9)+(3(9)+1)\omega) = \frac{9}{2}(54+28\omega) = 243+126\omega$. $X_{29} = (243+126\omega) + 28 - 29\omega^2 = 271+126\omega - 29(-1-\omega) = 271+126\omega+29+29\omega = 300+155\omega$. $Z = X_{29} - 155\omega = (300+155\omega) - 155\omega = 300$. Since $Z=300 \ge 0$, $n=29$ is a valid solution.

All three values $n=29, 30, 31$ satisfy the given condition. Since the question is presented as a single-choice question, and all three are options, there might be an implicit constraint not stated, or the question is flawed for a single-choice format. However, if forced to select one, and assuming the smallest possible integer value for $n$ is expected, then $n=29$.

The final answer is $\boxed{\text{29}}$.