How can it be true?

Algebra Level 2

${2^x + 2^{x+1} + 2^{x+2} + \ldots + 2^{x+2015} = 4^x + 4^{x+1} + 4^{x+2} + \ldots + 4^{x+2015}}$

If $x$ satisfies the equation above and it can be represented as $\log_D \left(\dfrac{A}{1+B^C} \right)$ for positive integers $A$ , $B$ , $C$ , and $D$ , where $B$ is prime, determine the smallest value of $A + B + C +D$ .

The answer is 2023.

1 solution

Chew-Seong Cheong
Aug 21, 2015

$\begin{aligned} 2^x + 2^{x+1} + 2^{x+2} + ... + 2^{x+2015} & = 4^x + 4^{x+1} + 4^{x+2} + ... + 4^{x+2015} \\ \frac{2^x (2^{2016} - 1 )}{2-1} & = \frac{4^x (4^{2016} - 1 )}{4-1} \\ 3\dot{} 2^x (2^{2016} - 1) & = 2^{2x} (2^{2016} - 1) (2^{2016} + 1) \\ 2^x (2^{2016} + 1) & = 3 \\ \Rightarrow x & = \log_2 \left( \frac{3}{1+2^{2016}} \right) \end{aligned}$

$\Rightarrow A+B+C+D = 3+2+2016+2 = \boxed{2023}$

This is the wrong answer.

Since it has asked for smallest value, 2^2016 can be written in many ways and for the required smallst value, it will be denoted as 64^336.

Then we get x=log(base-2)[3÷(1+64^336)]

Hence A=3 B=64 C=336 D= 2

And A+B+C+D= 3+64+336+2 =405

Raunak Agrawal - 4 years, 3 months ago

B is prime

Abhishek Maurya - 3 years, 3 months ago

did same...

Dev Sharma - 5 years, 5 months ago

Team plz respond .

Raunak Agrawal - 4 years, 3 months ago

B is prime

Jerry McKenzie - 4 years, 1 month ago

How can it be true?

The answer is 2023.

1 solution

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