[13^12 + 5(2^03^12+2^13^11+2^23^10++2^113) = 2^n m,] where m is odd.

Question

$$\frac{1}{3^{12}} + 5\left(\frac{2^0}{3^{12}}+\frac{2^1}{3^{11}}+\frac{2^2}{3^{10}}+\cdots+\frac{2^{11}}{3}\right) = 2^n\cdot m,$$
where $m$ is odd. Then $m\cdot n$ is equal to \_\_\_\_\_\_\_\_.

Answer: 12

Accepted Answer

The geometric part sums to $\dfrac{6^{12}-1}{3^{12}\cdot5}$; substituting:

$$\frac{1}{3^{12}} + \frac{6^{12}-1}{3^{12}} = \frac{6^{12}}{3^{12}} = 2^{12}.$$

So $2^n\cdot m = 2^{12}$, giving $m=1$, $n=12$, and $m\cdot n=12$.

Solutions · 1