Solve this Math Problem: Find Last 2/3 Digits of 3^3^3^3

[ilml]

Hello, I was randomly browsing sites and I came upon this math question which I am stuck on (and desperately want to solve!): Find the last two digits of 3^3^3^3. Can you find the last three digits?

I'm stuck on this problem because I really don't know where to begin. I figure that you need to find a pattern or period of some sort, but the numbers grow too fast and therefore there aren't that many readily available samples. Any help is appreciated! THANKS!

 

[arildno]

Instead of tackling the original problem, you should try first to find the answer to the following problem:
What is the last digit of the tower?
Note the following:
We have the recurrence pattern for the last digit:
3->3
3*3->9
3*3*3->7
3*3*3*3->1

And back again!
Hence, we have a "4-periodicity" for the last digit, and you should be able to solve the problem.

Clearly, you may grit your teeth together and find similar reccurrence patterns for the 2 last digits and 3 last digits problem.

However, I would believe that such recurrence patterns are well known in number theory; since I'm ignorant of that field, I can only suggest that you post your question in the "Number Theory" subforum.

 

[M98Ranger]

To solve this problem, we can use the concept of modular arithmetic. The last two digits of a number can be found by dividing the number by 100 and taking the remainder.

First, let's simplify the expression by finding the last two digits of 3^3 and 3^3^3.
3^3 = 27, so the last two digits are 27 % 100 = 27.
3^3^3 = 27^3 = 19683, so the last two digits are 19683 % 100 = 83.

Now, we can use this information to find the last two digits of 3^3^3^3.
3^3^3^3 = 3^19683.
We can use the concept of modular exponentiation to simplify this expression.
3^n % 100 = (3 % 100)^n % 100
So, we can rewrite 3^19683 as (3 % 100)^19683 % 100.
Now, we can substitute the values we found earlier for 3^3 and 3^3^3.
3^19683 = (27)^83 % 100.
Using the same method as before, we can simplify this to (27 % 100)^83 % 100 = 83^83 % 100.

To find the last two digits, we can use the cyclicity of digits.
The last two digits of powers of 83 follow a pattern of 83, 89, 47, 41, 43, 49, 87, 81, 83...
So, the last two digits of 83^83 will be the same as the last two digits of 83^(83 % 8) = 83^3 = 27.
Therefore, the last two digits of 3^3^3^3 are 27.

Similarly, to find the last three digits, we can use the same method but instead find the last three digits of 3^3^3^3.
3^3^3^3 = 3^19683.
Using the concept of modular exponentiation, we can rewrite this as (3 % 1000)^19683 % 1000.
Substituting the values we found earlier, we get (27 % 1000)^83 % 1000