First digit of powers of 2

Number Theory Level 3

Here are the first ten and second ten powers of two

$2^{0}=1$	$2^{10}=1024$
$2^{1}=2$	$2^{11}=2048$
$2^{2}=4$	$2^{12}=4096$
$2^{3}=8$	$2^{13}=8192$
$2^{4}=16$	$2^{14}=16384$
$2^{5}=32$	$2^{15}=32768$
$2^{6}=64$	$2^{16}=65536$
$2^{7}=128$	$2^{17}=131072$
$2^{8}=256$	$2^{18}=262144$
$2^{9}=512$	$2^{19}=524288$

Notice the first digit of each power of two is the same as the power 10 larger.

Is it true that for any $n \ge 0$ $\large 2^{n} \text{ and } 2^{n+10}$ will begin with the same digit?

No Yes

2 solutions

Jeremy Galvagni
Aug 13, 2018

The reason they usually begin with the same digit is $2^{10}$ is pretty close to $1000$ . If $2^{n}$ is close to being a higher digit, $2^{n+10}$ can be pushed over the top.

The first counterexample is $2^{36}=6.87 \times 10^{10}$ but $2^{46}=7.04 \times 10^{13}$

X X
Aug 15, 2018

$\log 2$ doesn't equal to $3$ ( $\log 2\approx0.3010$ ), so the answer is no

Could you explain why this fact (log 2 =/= 3) shows the answer is no?

Jason Carrier - 2 years, 10 months ago

Because $2^n=10^{n\log2}$ , so if $n$ become bigger, it will not follow the rule.

X X - 2 years, 9 months ago

First digit of powers of 2

2 solutions

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