We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
162
5
avatar

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
avatar+19687 
0

DELETED......

 Feb 23, 2019
edited by ElectricPavlov  Feb 23, 2019
 #2
avatar
0

????

Guest Feb 23, 2019
 #3
avatar
0

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
avatar+105448 
0

\(7^{7^7}\)   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
avatar
+1

tysm!

Guest Feb 23, 2019

36 Online Users

avatar
avatar
avatar
avatar