(Base Case) Let n = 25. Then 25! > 10^25. So the base case holds.
(Inductive Assumption) Now let n be an arbitrary number, such that n ≥ 25. Assume that n! > 10^n.
Since n ≥ 25, n+1 > 10. Also, since we already know n! > 10^n:
(n+1)*n! > 10*10^n, or rather, (n+1)! > 10^(n+1).
∴By the principle of mathematical induction, n! > 10^n for all n ≥ 25. QED.
]]>We know that for any number greater than 25, n!>10^n.
Is this a theorem or ? Thanks very much for your explanation. Happy holidays.
]]>2^10,000 contain 3011 digits.
The factorial of this number would be of the order of 10^10^3011 (actually, much greater!).
Sorry again for asking but can you explain how did you got to this - (2^10 000)! = 10^10^3011 I tried the Stirling formula but nothing like this Thanks again.
]]>By definition, log(2) 2^10000 = 10000
And by combining that with the rule that log(a)b = log(c)b/log(c)a, you can work it out.
log(2) 2^10000 = [log(10) 2^10000] ÷ [log(10) 2]
∴ log(10) 2^10000 = log(2) 2^10000 * log(10) 2 = 10000 * log(10)2 = 3010.3...
So rounding up gives that 2^10000 has 3011 digits.
]]>2^10,000 contain 3011 digits.
The factorial of this number would be of the order of 10^10^3011 (actually, much greater!).
This is true because for any number n greater than 100, n!>>10^n.
]]>lg(lgX)) > 2500
lgX > 10^2500
X > 10^(10^2500)
(2^10000)! > 10^(10^2500)
These are both extremely large numbers. Most calculators don't even like 80!, so (2^10000)! is huge! Similarly, the right-hand side will have (10^2500)+1 digits!
I can't see how to get the (2^10000)! into a usable form right now, but I'll be back later to ponder over it some more.
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