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.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

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