### 2^(3^(4^(5^(6^(7^(8^9)))))) - Part I

The problem comes from the IBM Ponder This October 2014 Challenge:

http://domino.research.ibm.com/Comm/wwwr_ponder.nsf/Challenges/October2014.html

This is a big number.

If you just take 8^9, you'll end up with 134,217,728. Now try raising, say, 2 to the power of this number, or 2^134,217,728. For comparison sake, the number of atoms in the universe is less than 2^324.

The problem is asking for the last 10 decimal digits, so that should have been an important clue. It should have rang a bell for two key ideas behind solving this problem:

1)

2)

Solution in the next post. Have fun!

http://domino.research.ibm.com/Comm/wwwr_ponder.nsf/Challenges/October2014.html

**What are the last 10 decimal digits of 2^(3^(4^(5^(6^(7^(8^9))))))?**This is a big number.

If you just take 8^9, you'll end up with 134,217,728. Now try raising, say, 2 to the power of this number, or 2^134,217,728. For comparison sake, the number of atoms in the universe is less than 2^324.

The problem is asking for the last 10 decimal digits, so that should have been an important clue. It should have rang a bell for two key ideas behind solving this problem:

1)

*Modular Arithmetic*: to get the last 3 digits of 1234, all I have to do is calculate 1234%(10^3)2)

*Big Integers*. Try Binging for Big Integers in C#.Solution in the next post. Have fun!

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