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

Whoops XD

That seems overly complex

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

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

(integers.Where(x => x % 2 == 0).Count() > 1 ? 1 : 0)
This bit will give a 1 if the number you are looking for is odd and 0 if you are looking for an even number (from asking if there are more than 1 even number with)
Using this we can find the 1 number in the array that matches this