Compare which of the following is bigger:
101611⋅301631 versus 201642
My solution:
I know there are multiple approaches by which we can use to tackle the problem at hand. But if it were up to me, I would first look for the prime factorization for each number involved in the problem, and I found:
1016=23⋅127
2016=25⋅32⋅7
3016=23⋅13⋅29
With the help of Wolfram Alpha, I've found
101611⋅301631>201642
The effort remains now to first assume 101611⋅301631>201642 is true and proved it so.
That is,
(23⋅127)11⋅(23⋅13⋅29)31>(25⋅32⋅7)42
Simplify the above we get:
233⋅12711⋅293⋅1331⋅2931>2210⋅384⋅742
12711⋅1331⋅2931>2210−33−93⋅384⋅742
12711⋅1331⋅2931>284⋅384⋅742
\displaystyle \color{yellow}\bbox[5px,purple]{127^{11}\cdot 377^{31}\gt 6^{84}\cdot 7^{42}}*
Here comes the difficult part of the process, we have to use the calculator to do a series of checking to see if we can get a simple relation between 377 and 7^a for some positive integer a such that 377\gt 7^a.
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}
We have made some good progress, as we have kicked out the 7^{42} in the RHS of the equation * and also reduced the power of 377 from 31 down to 31-14=17.
At this point, if we can prove
\displaystyle \color{yellow}\bbox[5px,green]{127^{11}\cdot 377^{17}\gt 6^{84}}
Then we can conclude \displaystyle 1016^{11}\cdot 3016^{31}\gt 2016^{42} is correct.
Again, we have to find a relation between 127\cdot 377 and 6^b for some positive integer b such that 127\cdot 377\gt 6^b.
Observe 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 \color{black}\bbox[5px,orange]{127^{14}\cdot 377^{14}\gt 6^{84}}
It's obvious that
\displaystyle \color{yellow}\bbox[5px,blue]{127^{11}\cdot 377^{17}\gt 127^{14}\cdot 377^{14} } is true, therefore, we've proved that
\displaystyle 1016^{11}\cdot 3016^{31} is bigger than \displaystyle 2016^{42}.
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