RegEx: How can I match all numbers greater than 49?

Try a conditional group matching 50-99 or any string of three or more digits:

var r = /^(?:[5-9]\d|\d{3,})$/

The fact that the first digit has to be in the range 5-9 only applies in case of two digits. So, check for that in the case of 2 digits, and allow any more digits directly:

^([5-9]\d|\d{3,})$

This regexp has beginning/ending anchors to make sure you're checking all digits, and the string actually represents a number. The | means "or", so either [5-9]\d or any number with 3 or more digits. \d is simply a shortcut for [0-9].

Edit: To disallow numbers like 001:

^([5-9]\d|[1-9]\d{2,})$

This forces the first digit to be not a zero in the case of 3 or more digits.

I know there is already a good answer posted, but it won't allow leading zeros. And I don't have enough reputation to leave a comment, so... Here's my solution allowing leading zeros:

First I match the numbers 50 through 99 (with possible leading zeros):

0*[5-9]\d

Then match numbers of 100 and above (also with leading zeros):

0*[1-9]\d{2,}

Add them together with an "or" and wrap it up to match the whole sentence:

^0*([1-9]\d{2,}|[5-9]\d)$

That's it!

Next matches all greater or equal to 11100:

^([1-9][1-9][1-9]\d{2}\d*|[1-9][2-9]\d{3}\d*|[2-9]\d{4}\d*|\d{6}\d*)$

For greater or equal 50:

^([5-9]\d{1}\d*|\d{3}\d*)$

See pattern and modify to any number. Also it would be great to find some recursive forward/backward operators for large numbers.