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!

Guest Feb 23, 2019

#3**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**

Guest Feb 23, 2019

#4**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.

Melody Feb 23, 2019