# Python Challenge - Level 0

## Problem

Hint: try to change the URL address.

## Explore

As the "0" indicates, this is just a warm up.

### Loop

The naive way to solve it: multiply by 2 in a loop:

>>> k = 1
>>> for i in range(38):
...     k *= 2
...
>>> k
274877906944

### Power

Use ** for power:

>>> 2**38
274877906944

In REPL the result will be printed; or you can explicitly print the result by calling print()

>>> print(2**38)
274877906944

Instead of **, you can also use pow()

>>> pow(2,38)
274877906944

or

>>> print(pow(2,38))
274877906944

### Shift

Multiply by 2 is equivalent to shifting the binary representation left by one:

>>> 1<<38
274877906944

### Numeric Types

Done!

Wait, why 38? what is implied?

If you are coming from C/C++, Java or other languages, you know that there are multiple types just for integers: short, integer, long, and even BigInteger beyond 64-bit. However that is not the case in Python:

>>> type(2**3)
<class 'int'>
>>> type(2**38)
<class 'int'>
>>> type(2**380)
<class 'int'>

So 38 is a good(and random) example to show a int larger than 32-bit.

Similar for float type, in Python it can be arbitrarily large, no double needed

>>> type(2**3.8)
<class 'float'>
>>> type(2.0**38)
<class 'float'>

## Complete Solution

print(2**38)

Output:

274877906944

## Next Level

http://www.pythonchallenge.com/pc/def/274877906944.html