Because it is. Framework is there to use it. :)

Just saying, I'm aware of BigInteger and there's no reason not to use it - just never done something like this before 👍

Please raise your code problems as a question and not an issue ;-)

Your code has bug.
Check your comments again please ;)

Whoops XD

Ok. Thanks for the comment.
Personally I don't like to nest commands inside commands. After some months, I'm sure I will forget the logic I have used :). Sometimes readability rewards you in a long term.

Your code will execute more linqs than mine :).

That seems overly complex

Thanks for your feedback. I am a Ruby developer and i was a .net developer just trying to refresh my .net skills. Will change as per your comment.

As you are using Linq, you can use it to output the number in a single line without needing to create lists, you can find out if the number you are looking for is odd or even (integers.Where(x => x % 2 == 0).Count() > 1 ? 1 : 0) so if this is 1 then we look for odd and 0 for even, we can then find where a variable fits the criteria integers.Where(z => (z % 2 == (*that code snippit before*))) and get the only value .First();, so now we have odd if the only number is odd and even if the only number is even.

Great start with Linq

As you are using Linq, you can use it to output the number in a single line without needing to create lists, you can find out if the number you are looking for is odd or even (integers.Where(x => x % 2 == 0).Count() > 1 ? 1 : 0) so if this is 1 then we look for odd and 0 for even, we can then find where a variable fits the criteria integers.Where(z => (z % 2 == (*that code snippit before*))) and get the only value .First();, so now we have odd if the only number is odd and even if the only number is even.

Great start with Linq

As you are using Linq, you can use it to output the number in a single line without needing to create lists, you can find out if the number you are looking for is odd or even (integers.Where(x => x % 2 == 0).Count() > 1 ? 1 : 0) so if this is 1 then we look for odd and 0 for even, we can then find where a variable fits the criteria integers.Where(z => (z % 2 == (*that code snippit before*))) and get the only value .First();, so now we have odd if the only number is odd and even if the only number is even.

Yours is way clearer though :P

As you are using Linq, you can use it to output the number in a single line without needing to create lists, you can find out if the number you are looking for is odd or even (integers.Where(x => x % 2 == 0).Count() > 1 ? 1 : 0) so if this is 1 then we look for odd and 0 for even, we can then find where a variable fits the criteria integers.Where(z => (z % 2 == (*that code snippit before*))) and get the only value .First();, so now we have odd if the only number is odd and even if the only number is even.

Great start with Linq