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Which is greater value in 50^99 and 99!. Solution with AM, GM inequality, and by Direct Multiplication.
Solution 1:
As we know that AM ≥ GM:
For 1 to 99,
\[\frac{1+2+....+98+99}{99}\geq \sqrt[99]{1\cdot2\cdot....\cdot98\cdot99}\]
\[\Rightarrow\frac{\frac{(99)(100)}{2}}{99}\geq \sqrt[99]{1\cdot2\cdot....\cdot98\cdot99}\]
\[\Rightarrow\frac{100}{2}\geq \sqrt[99]{1\cdot2\cdot....\cdot98\cdot99}\]
\[\Rightarrow 50\geq \sqrt[99]{99!}\]
\[\Rightarrow 50^{99}\geq 99!\]
It is clear that,
\[50^{99} \neq 99!\]
(50 contains only 2s and 5s but 99! contain all prims up to 99)
\[\Rightarrow 50^{99}>99!\]
Solution 2:
For 99! we will create groups as (1·99), (2·98), etc. up to (49·51)
There are a total of 49 pairs and one term is 50.
We know that,
\[50\cdot50>1\cdot99\]
\[50\cdot50>2\cdot98\]
.
.
.
\[50\cdot50>48\cdot52\]
\[50\cdot50>49\cdot51\]
And in the last, 50=50
By multiplying all of the above values, we can say that:
\[50^{99}>99!\]
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