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# Base 7 Question

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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
+18942
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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
+103689
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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
+1

tysm!

Guest Feb 23, 2019