Find how many digits in the base 10 number 9^(9^9) has when written in base nine

Hans007
Feb 11, 2018

#6**+1 **

**Guest, Your answer does not contradicted mine :)**

This was the question:

**Find how many digits in the base 10 number 9^(9^9) has when written in base nine**

Your comment was

So, the number of digits =369,693,099 + 1 =369,693,100 digits - in base 10

YES I agree,

This whole thing is in base 10

What I said is that the answer is

it will have 9^9+1 digits = 387 420 490 digits (base 10) BUT this is for the base 9 answer.

To put it slightly differently:

The base 9 conversion of 9^(9^9) will have 387 420 490 digits, with the number of digits expressed in base 10,

If the number of digits was expressed in base 9 it would look bigger.

I can do the conversion if you want:

387420490/9 = 43 046 721 R1

43046721/9= 4 782 969 R0

4782969/9 = 531 441 R0

531441/9 = 59049 R0

59049/9 = 6561 R0

6561/9 = 729 R0

729/9 = 81 R0

81/9 = 9 R0

9/9 = 1 R0

1/9 = 0 R1

**So the number of digits base 9 is 1 000 000 001**

This certainly is no surprise I expected it to be a 1 at the beginning and at the end with zeros in the middle.

10^(10^10) base 10 = 10^10 000 000 000 = 1 followed by 10 000 000 000 zerso = 10 000 000 001 digits

**I just discovered the pattern ...**

2^(2^2) = 2^4 = 10000(base2) : 5 digits long base 10 and 5 base 10 = 101 base 2

3^(3^3)= 3^27 = 1 followed by 27 zeros in base 3 : 28 digits long base 10 which is 1001 base 3

4^(4^4) = 4^ 256 = 1 followed by 256 zeros base 4: 257 digits long base 10 which is 10001 base 4

...

by extension:

**9^(9^9) base 10 will be 1 000 000 001 base 9 units long in base 9. Which is exactly what I got when I did it the super long way!!!**

Melody
Feb 11, 2018

#2**0 **

Find how many digits in the base 10 number 9^(9^9) has when written in base nine

it will have 9^9+1 digits = 387 420 490 digits (base 10)

-------------

simple example ( I needed this to think through my answer)

2^(2^2) base 10 written in base 2

2^4=16=10000_{2}

So that is 2^2+1=5 digits base 10

OR 5_{10 }= 101_{2}

Melody
Feb 11, 2018

#3**+1 **

Melody: 9^(9^9) =9^(387,420,489) =387,420,489 x log(9) =369,693,099.631........etc.

So, the number of digits =369,693,099 + 1 =369,693,100 digits - in base 10

See Wolfram/Alpha Here:

http://www.wolframalpha.com/input/?i=9%5E9%5E9

Guest Feb 11, 2018

edited by
Guest
Feb 11, 2018

#6**+1 **

Best Answer

**Guest, Your answer does not contradicted mine :)**

This was the question:

**Find how many digits in the base 10 number 9^(9^9) has when written in base nine**

Your comment was

So, the number of digits =369,693,099 + 1 =369,693,100 digits - in base 10

YES I agree,

This whole thing is in base 10

What I said is that the answer is

it will have 9^9+1 digits = 387 420 490 digits (base 10) BUT this is for the base 9 answer.

To put it slightly differently:

The base 9 conversion of 9^(9^9) will have 387 420 490 digits, with the number of digits expressed in base 10,

If the number of digits was expressed in base 9 it would look bigger.

I can do the conversion if you want:

387420490/9 = 43 046 721 R1

43046721/9= 4 782 969 R0

4782969/9 = 531 441 R0

531441/9 = 59049 R0

59049/9 = 6561 R0

6561/9 = 729 R0

729/9 = 81 R0

81/9 = 9 R0

9/9 = 1 R0

1/9 = 0 R1

**So the number of digits base 9 is 1 000 000 001**

This certainly is no surprise I expected it to be a 1 at the beginning and at the end with zeros in the middle.

10^(10^10) base 10 = 10^10 000 000 000 = 1 followed by 10 000 000 000 zerso = 10 000 000 001 digits

**I just discovered the pattern ...**

2^(2^2) = 2^4 = 10000(base2) : 5 digits long base 10 and 5 base 10 = 101 base 2

3^(3^3)= 3^27 = 1 followed by 27 zeros in base 3 : 28 digits long base 10 which is 1001 base 3

4^(4^4) = 4^ 256 = 1 followed by 256 zeros base 4: 257 digits long base 10 which is 10001 base 4

...

by extension:

**9^(9^9) base 10 will be 1 000 000 001 base 9 units long in base 9. Which is exactly what I got when I did it the super long way!!!**

Melody
Feb 11, 2018