6th grade problem

Algebra Level 2

$\begin{aligned} \large 1 + a + a^2 + a^3 + \ldots + a^x &=& (1+a)(1+a^2)(1+a^4)(1+a^8) \\ \large x &=& \ ? \end{aligned}$

Assume $a \ne 0$ .

The answer is 15.

2 solutions

Patrick Engelmann
Apr 28, 2015

$1 + a^{} + a^{2} + a^{3} + \dots + a^{x} = \frac{1-a^{x+1}}{1-a}$

$\Rightarrow \frac{1-a^{x+1}}{1-a} = (1+a)(1+a^{2})(1+a^{4})(1 + a^{8})$

$\Leftrightarrow 1-a^{x+1} =(1+a)(1-a)(1+a^{2})(1+a^{4})(1 + a^{8})$

$\Leftrightarrow 1-a^{x+1} =(1-a^{2})(1+a^{2})(1+a^{4})(1 + a^{8})$

$\Leftrightarrow 1-a^{x+1} =(1-a^{4})(1+a^{4})(1 + a^{8})$

$\Leftrightarrow 1-a^{x+1} =(1-a^{8})(1 + a^{8})$

$\Leftrightarrow 1-a^{x+1} = 1-a^{16}$

$\Leftrightarrow x = \boxed{15}$

Curtis Clement
Apr 19, 2015

You can expand both sides are cancel out like terms, but there is a shortcut... For the identity to always be true both sides must contain the same highest power which is: $8+4+2+1 = \boxed{15}$

Moderator note:

Of course expanding and canceling out like terms works. There's a simple solution to this.

Your shortcut is not sound, how do you guarantee that there are all fifteen terms $1,a,a^2,\ldots,a^{14}$ as well?

6th grade problem

The answer is 15.

2 solutions

Moderator note:

0 pending reports