# True or False: $2^{2^{2011}} \text{ divides } 2^{2^{2012} }$

True or false: $$2^{2^{2011}} \text{ divides } 2^{2^{2012} }$$

Give your justifications. I don't know how to start this problem so far.

But, I guessed like this, $$2^{\underbrace{2\times 2\times\ldots\times 2}_{2011 times}}\text{ divides } 2^{\underbrace{2\times 2\times\ldots\times 2}_{2012 times}}$$

$$\implies \qquad \underbrace{2\times2\times\ldots\times 2}_{2^{2011}\;times}\text{ divides } \underbrace{2\times2\times\ldots\times 2}_{2^{2012}\;times}$$

Number of factors in the first product is less than that of second product. So defenitly,it will divide.

Kindly say is my reason is sound? If its ok, then give me alternate details about if we change the base . $\left(\text{That is what about }a^{2^{2011}}\right.$ and $\left.b^{2^{2012}}\right)$ where $gcd(a,b)=d\neq 1$

• Doesn't it follow directly from the fact that $2^{2012} > 2^{2011}$ ? Jan 16, 2015 at 14:08
• Give me alternate details..please.. I dont want the shorter version of my proof. But, different approach.. Jan 16, 2015 at 14:31

The reasoning here is very simple. $2^n$ divides $2^m$ for $n,m$ positive integers if and only if $n \le m$ ; this is just because you multiply a bunch of $2$'s together, so if there are less multiples of $2$ on one side than on the other, then the side with less $2$'s divides the other side.

Now $n=2^{2011} \le 2^{2012}=m$ shouldn't be hard to see.

Hope that helps,

• yah.. I also used the same thing in my proof. Can you give suggestion about last quote?@Patrick Jan 16, 2015 at 14:24
• Try proving that $(a^n,b^m) = (a,b)^{\min \{n,m \}}$, where $(a,b)$ denotes the g.c.d. of $a$ and $b$. Jan 17, 2015 at 18:42

More specifically, $2^{2^{2012}}=2^{2^{2011}\times2}=(2^{2^{2011}})^{2}$

• Indeed, this is the simplest proof IMO. Jan 16, 2015 at 15:38

True because $$2011 +1 = 2012 \Longrightarrow 2^{2012} = 2 \cdot 2^{2011} \Longrightarrow 2^{2^{2012}}= (2^{2^{2011}})^2$$