# How many factors $2$ the number $3^{x} + 1$ has given that $x$ is an even number?

First of all, given that $x$ is a whole even number, let's write $x = 2m$ for some natural $m$. Then observe that we are allowed, according to Newton, to write

$3^{2m} = (1+2)^{2m} = \sum_{n=0}^{2m} \begin{pmatrix} 2m\\ n \end{pmatrix} 2^{n} = 1 + \sum_{n=1}^{2m} \begin{pmatrix} 2m\\ n \end{pmatrix} 2^{n}$

And notice that

$\sum_{n=1}^{2m} \begin{pmatrix} 2m\\ n \end{pmatrix} 2^{n}$

is an even number, since the factor $2$ happens to appear in all of the terms (and this is only possible because $0 < n < 2m + 1$).

Now, this is $3^{x}$. Adding the $1$ we get

$3^{2m} + 1 = 2 + \sum_{n=1}^{2m} \begin{pmatrix} 2m\\ n \end{pmatrix} 2^{n} = 2 \times \left [ 1 + \sum_{n=1}^{2m} \begin{pmatrix} 2m\\ n \end{pmatrix} 2^{n-1} \right ]$

But

$1 + \sum_{n=1}^{2m} \begin{pmatrix} 2m\\ n \end{pmatrix} 2^{n-1}$

is an odd number, since

$\begin{pmatrix} 2m\\ 1 \end{pmatrix}$

is even (for $n = 1$, where $2^{n-1} = 1$) and for all the other terms $2^{n-1}$ is even. Therefore

$1 + \sum_{n=1}^{2m} \begin{pmatrix} 2m\\ n \end{pmatrix} 2^{n-1}$

has no factors $2$ and

$2 \times \left [ 1 + \sum_{n=1}^{2m} \begin{pmatrix} 2m\\ n \end{pmatrix} 2^{n-1} \right ]$

has all the factors $2$ in evidence. Then $3^{x} + 1$ for an even $x$ has only one factor $2$. Note by Lucas Tell Marchi
6 years, 5 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

Great!

There is a one line solution to this problem, with a similar approach to what you did. Can you figure that out?

Staff - 6 years, 5 months ago

Haven't found it yet :P

- 6 years, 5 months ago

Hint: $9 = 1 + 8$

Staff - 6 years, 5 months ago

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

for some natural $a$

- 6 years, 5 months ago

Perfecto! Great job!

In fact, it is $2 ( 1 + 4b )$.

Staff - 6 years, 5 months ago