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Find how many digits the base 7 number \({7}^{({7}^{7})}\) has when written in base 7.

 

I tried using a "big number calculator", but the answer was several pages long. Obvisously, there's a better way to do this.

 

Thanks!

 Feb 23, 2019
 #1
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DELETED......

 Feb 23, 2019
edited by ElectricPavlov  Feb 23, 2019
 #2
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????

Guest Feb 23, 2019
 #3
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7^(7^7) =3.759823527 E+695,974 in base 10

 

Convert 7^(7^7) from base 10 to base 7 =6.666666666666..._7×7^823,542

 Feb 23, 2019
 #4
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\(7^{7^7}\)   is not a base 7 number  (there are no 7s in a base 7 number)     so I am guessing this is in base 10

 

Anyway, when it is written in full in base 7 it will have the same number of digits as     \(10^{10^{10}}\)    (base 10) has

 

\({10^{10}} \)    has 10 zeros plus the 1 which is 11 digits  = 10 000 000 000

 

\(10^{10000000000}\)     has 10 billion zeros plus a 1 in the biggest spot

 

So altogether  \(10^{(10^{10})}\)   will have      10,000,000,001     digits.

 

so

 

\(7^{(7^7)} \;\;base\;10\\ =10^{(10^{10})} \;\;base\;7\\ =\text{1 followed by 10,000,000,000 zeros}\;\;\\ =\text{10,000,000,001 places}\)

 

I think that is correct but I always get confused when dealing with huge numbers so it would be good if someone could confirm it.

 Feb 23, 2019
 #5
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tysm!

Guest Feb 23, 2019

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