What are the first and the last 20 digits of this very large number: 13^(13^13) =13^302,875,106,592,253 ?

Any help would be greatly appreciated. Thank you.

Guest Oct 4, 2019

#1**+1 **

First Part - The first 20 digits. This is very simple.

13^13 = 302,875,106,592,253

Multiply the above number by log 13(base 10). Carry the result to about 100 decimal places. You should get:

302,875,106,592,253 x log 13 =337,385,711,567, 664.8232318987826741897646811857360282268728608288726576980008833486046624631780203460135940132237440066.

From the above result, keep the fractional part ONLY.

Raise the fractional to the power 10:

10^0.8232318987826741897646811857360282268728608288726576980008833486046624631780203460135940132237440066 =6.656284847586597803523004328401344362443297464463201858579506080321635285934050922391456377579985062630734182573941. Remove the decimal point and count the first 20 digits from the left **=66,562,848,475,865,978,035 - This is the first 20 digits of your number.**

Second Part - The last 20 digits.

This part is very difficult to do without a programming language. For this purpose, I wrote a short code in Python which gives you the last 20 digits:

1- %%time

2 - # Fast python code that first calls pow()

3 - # then applies % operator

4 - import math as m

5 - a =13

6 - b =302875106592253

7 - p =100000000000000000000

8 - # Using direct fast method to compute

9 - # (a ^ b) % p.

10 - d = pow(a, b, p)

11 - print( f"{d :,d}")

**The last 20 digits are = 91,055,086,363,688,549,053**

Guest Oct 4, 2019