Comparison between two numbers

[anemone]

Determine, without the help of calculator, which of the following is bigger:

\(\displaystyle 1016^{11}\cdot 3016^{31}\) versus \(\displaystyle 2016^{42}\)

 

[kaliprasad]

Spoiler

We know $3^6(= 729) > 2^9(=512)$ hence $3^{30} > 2^{45}$
Or $3^{31} > 3 * 2^{45} > 24 * 2^{42}$
so $(1000)^{11} * {3000}^{31} > 24 * {2000}^{42}$
Hence $(1016)^{11} * {3016}^{31} > 24 * {2000}^{42}\cdots(1)$
Now $2016^{42} = 2000^{42}( 1+ \frac{1}{125})^{42} < 2000^{42}( 1+ \frac{1}{125})^{125} < 2000^{42} * e < 2000^{42} *3$
from above and (1) we have $(1016)^{11} * {3016}^{31} > {2016}^{42}\cdots(3)$
Hence $(1016)^{11} * {3016}^{31}$ is larger

 

[anemone]

Awesome! Thanks kaliprasad for participating!(Cool)

My solution:

Spoiler

I first assume \(\displaystyle 1016^{11}\cdot 3016^{31}\gt 2016^{42}\).

From the prime factorization of \(\displaystyle 1016=2^3\cdot 127,\,2016=2^5\cdot 3^2\cdot 7,\,3016=2^3\cdot 13\cdot 29\)

I need to prove

\(\displaystyle (2^3\cdot 127)^{11}\cdot (2^3\cdot 13\cdot 29)^{31}\gt (2^5\cdot 3^2\cdot 7)^{42}\)

Simplify the above we get:

\(\displaystyle 2^{33}\cdot 127^{11}\cdot 2^{93}\cdot 13^{31}\cdot 29^{31}\gt 2^{210}\cdot 3^{84}\cdot 7^{42}\)

\(\displaystyle 127^{11}\cdot 13^{31}\cdot 29^{31}\gt 2^{210-33-93}\cdot 3^{84}\cdot 7^{42}\)

\(\displaystyle 127^{11}\cdot 13^{31}\cdot 29^{31}\gt 2^{84}\cdot 3^{84}\cdot 7^{42}\)

\(\displaystyle 127^{11}\cdot 377^{31}\gt 6^{84}\cdot 7^{42}\)*

Observe that:

$7^2=49,\,7^3=343\implies 373\gt 7^3$

Therefore $373^{14}\gt (7^3)^{14}$ which is $373^{14}\gt 7^{42}$

At this point, if we can prove

\(\displaystyle 127^{11}\cdot 377^{17}\gt 6^{84}\)

Then we can conclude \(\displaystyle 1016^{11}\cdot 3016^{31}\gt 2016^{42}\) is correct.

Observe again that

$127\cdot 377=47879\gt 6^6=46656$

Taking 14th power on both sides of the inequality we get:

$(127\cdot 377)^{14}\gt (6^6)^{14}$

\(\displaystyle 127^{14}\cdot 377^{14}\gt 6^{84}\)

It's obvious that

\(\displaystyle 127^{11}\cdot 377^{17}\gt 127^{14}\cdot 377^{14}\) is true, therefore, our assumption has been justified, therefore \(\displaystyle 1016^{11}\cdot 3016^{31}\) is bigger than \(\displaystyle 2016^{42}\).