Function to view N-th digit of large number?

Questions
1

Hello,
I would like to view numbers in the string on wanted position, eg. pi, e, 9^(9^9),…

I would like to see the digit on position 177486336 of the number 9^(9^9).
I read a lot on internet but I thing I missed something.

Thank you for recommendation.

2

I don’t know if this is the best way to do this, especially for big numbers, but my attempt gave this result:

a = 9^(9^9);
Take[IntegerDigits[a], {177486336}]

(* 3 *)

4268im1.png
4268im1.png316×100 1.64 KB

If the precision of this command for large numbers is correct this should be the result…

Just for curiosity this is the amount of digits of this number:

IntegerLength[9^(9^9)]

(* 369693100 *)

3

This looks great. I tried but it gave me this:

Beznázvu.png
Beznázvu.png487×273 8.99 KB

I know the length of this number and i would like to see

  • 7-digit sequence at position 177486336
  • 8-digit sequence at position 352695442
4

I got these results:

a = 9^(9^9);
{FromDigits[Take[IntegerDigits[a], {177486336, 177486342}]]}
{FromDigits[Take[IntegerDigits[a], {352695442, 352695449}]]}

5750im2.png
5750im2.png454×154 2.84 KB

7 digits starting at the position 177486336: (* 3955388 *)

8 digits starting at the position 352695442: (* 11623480 *)

5

Thank you.

6

Hello. I need to find the last digit of number 9^9^9, i tried

IntegerLength[9^(9^9)]

a i get a number of digits, but i cannot use

ResourceFunction["NthDigit"][9^(9^9),369693100]

or

ResourceFunction["NthDigit"][9^(9^9),{369693100}]

It didn´t work.
Please help me, how can i do it?

7

There is no need to compute all the digits to determine the last one. Observe the pattern

Table[9^i, {i, 1, 10, 2}]
(* {9, 729, 59049, 4782969, 387420489} *)

Table[9^i, {i, 2, 10, 2}]
(* {81, 6561, 531441, 43046721, 3486784401} *)

Odd powers end in 9, even powers end in 1. Since 9^9 is odd, the last digit of 9^(9^9) must be 9.

Verify by computing the number

9^(9^9) // IntegerDigits // Last
(* 9 *)