They Are Spelled The Same, Aren't They?

Number Theory Level 3

Which is larger? 2 30 ! or ( 2 30 ) ! \large 2^{30!} \quad \text{or} \quad (2^{30})!

2 30 ! { 2 }^{ 30! } ( 2 30 ) ! ({ 2 }^{ 30 })! They are equal

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Mohammad Hamdar by Mohammad Hamdar , 22, Lebanon

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1 solution

Discussions for this problem are now closed

First, note that

( 2 30 ) ! = 1 ⋅ 2 ⋯ 2 30 < 2 30 ⋅ 2 30 ⋯ 2 30 = 2 30 ⋅ 2 30 . (2^{30})! = 1 \cdot 2 \cdots 2^{30} < 2^{30} \cdot 2^{30} \cdots 2^{30} = 2^{30 \cdot 2^{30}}.

Then, using the fact that 2 30 < 29 ! 2^{30} < 29! (see comments for an explanation of this), 2 30 < 29 ! 2^{30} < 29! so 30 ⋅ 2 30 < 30 ! , 30\cdot 2^{30} < 30!, and thus 2 30 ⋅ 2 30 < 2 30 ! . 2^{30\cdot 2^{30}} < 2^{30!}.

Finally, this gives ( 2 30 ) ! < 2 30 ∗ 2 30 < 2 30 ! , (2^{30})! < 2^{30*2^{30}} < 2^{30!}, so ( 2 30 ) ! < 2 30 ! . (2^{30})! < 2^{30!}.

100 Helpful 0 Interesting 1 Brilliant 0 Confused

Nontrivial that 2 30 < 29 ! 2^{30} < 29!

Kane Williams - 5 years, 1 month ago

29 ! = 29 ⋅ 28 ⋅ … ⋅ 2 ⋅ 1 > 8 ⋅ ( 2 ⋅ 2 ⋅ … ⋅ 2 ) ⋅ 1 = 2 3 ⋅ 2 27 = 2 30 29! = 29 \cdot 28 \cdot \ldots \cdot 2 \cdot 1 > 8 \cdot (2 \cdot 2 \cdot \ldots \cdot 2) \cdot 1 = 2^3 \cdot 2^{27} = 2^{30}

Ivan Koswara - 5 years, 1 month ago

29 ! = 2 ∗ 3 ∗ . . . ∗ 29 29 ! = ( 2 ∗ 4 ∗ . . . ∗ 28 ) ( 3 ∗ 5 ∗ . . . ∗ 29 ) 29 ! = 2 14 ( 2 ∗ 3 ∗ . . . ∗ 14 ) ( 3 ∗ 5 ∗ . . . ∗ 29 ) 29 ! = 2 14 ( 2 ∗ 4 ∗ . . . ∗ 14 ) ( 3 ∗ 5 ∗ . . . ∗ 13 ) ( 3 ∗ 5 ∗ . . . ∗ 29 ) 29 ! = 2 21 ( 2 ∗ 3 ∗ . . . ∗ 7 ) ( 3 ∗ 5 ∗ . . . ∗ 13 ) ( 3 ∗ 5 ∗ . . . ∗ 29 ) 29 ! = 2 21 ( 2 ∗ 4 ∗ 6 ) ( 3 ∗ 5 ∗ 7 ) ( 3 ∗ 5 ∗ . . . ∗ 13 ) ( 3 ∗ 5 ∗ . . . ∗ 29 ) 29 ! = 2 24 ( 2 ∗ 3 ) ( 3 ∗ 5 ∗ 7 ) ( 3 ∗ 5 ∗ . . . ∗ 13 ) ( 3 ∗ 5 ∗ . . . ∗ 29 ) 29 ! = 2 25 ( 3 ) ( 3 ∗ 5 ∗ 7 ) ( 3 ∗ 5 ∗ . . . ∗ 13 ) ( 3 ∗ 5 ∗ . . . ∗ 29 ) 29 ! = 2 25 ∗ 3 4 ( 5 ∗ 7 ) ( 5 ∗ . . . ∗ 13 ) ( 5 ∗ . . . ∗ 29 ) 29! = 2 * 3 *...* 29 \\ 29! = (2 * 4 *...* 28)(3*5*...*29) \\ 29! = 2^{14}(2*3*...*14)(3*5*...*29) \\ 29! = 2^{14}(2*4*...*14)(3*5*...*13)(3*5*...*29) \\ 29! = 2^{21}(2*3*...*7)(3*5*...*13)(3*5*...*29) \\ 29! = 2^{21}(2*4*6)(3*5*7)(3*5*...*13)(3*5*...*29) \\ 29! = 2^{24}(2*3)(3*5*7)(3*5*...*13)(3*5*...*29) \\ 29!= 2^{25}(3)(3*5*7)(3*5*...*13)(3*5*...*29) \\ 29! = 2^{25}*3^{4}(5*7)(5*...*13)(5*...*29) \\

2 30 = 2 25 ∗ 2 5 2^{30} = 2^{25}*2^{5} \\

32 < 81 2 5 < 3 4 2 5 < 3 4 ( 5 ∗ 7 ) ( 5 ∗ . . . ∗ 13 ) ( 5 ∗ . . . ∗ 29 ) 2 30 < 2 25 ∗ 3 4 ( 5 ∗ 7 ) ( 5 ∗ . . . ∗ 13 ) ( 5 ∗ . . . ∗ 29 ) 2 30 < 29 ! 32 < 81 \\ 2^{5} < 3^{4} \\ 2^{5} < 3^{4}(5*7)(5*...*13)(5*...*29) \\ 2^{30} < 2^{25}*3^{4}(5*7)(5*...*13)(5*...*29) \\ 2^{30} < 29!

Reasonably nontrivial, I suppose...

Alexander Suarez-Beard - 5 years, 1 month ago

I like the way you proved. Comparing their logarithm of base e is however another practical way using calculator.

Lu Chee Ket - 4 years, 11 months ago

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