Comparing 2013! and 1007^{2013}.

frobirrimupyx 2022-09-13 Answered
Comparing 2013! and 1007 2013
I have to compare the following two numbers:
2013 !  and  1007 2013
where n ! = 1 × 2 × × ( n 1 ) × n
I tried in different ways to group the 1 × 2 × × 2012 × 2013 to obtain some kind of association with the 1007 from 1007 2013 but no luck.
Is there any standard approach for this kind of problem?
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Answers (2)

gasskadeu7
Answered 2022-09-14 Author has 21 answers
Step 1
Use Stirling’s approximation for the factorial:
n ! 2 π n ( n e ) n ,
so 2013 ! 112.46356 ( 740.54132 ) 2013 < 1007 2013 .
Step 2
For a less computational approach, look at their ratio:
2013 ! 1007 2013 = 2013 2012 2 1 1007 1007 1007 1007 = 2013 1007 2012 1007 2011 1007 1008 1007 1006  factors 1007 1007 1006 1007 2 1007 1 1007 1006  factors .
The 1006 factors on the left are between 1 and 2; the factors on the right are between 1 and 1 1007 . Try matching them up.

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rustenig
Answered 2022-09-15 Author has 1 answers
Step 1
1 × 2 × × 2012 × 2013
= ( 1007 1006 ) × ( 1007 1005 ) × × ( 1007 + 1005 ) × ( 1007 + 1006 )
= ( 1007 2 1006 2 ) × ( 1007 2 1005 2 ) × × ( 1007 2 1 2 ) × 1007
Step 2
< ( 1007 2 ) 1006 × 1007
= 1007 2013

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