Sum of digits in a number is Even or Odd?

Using While loop

DECLARE @num varchar(70), @len int, @sum int;

set @num = '9075453574568458765456786586578567856785677548964154022106422' + '0'

set @len = LEN(@num )

set @sum = 0

 while(@len>-1)

   begin

      set @sum = @sum + convert (int, RIGHT(@num ,1))

      set   @num = left(@num ,@len)

      set   @len -= 1

   end

select (case @sum%2 when 1  then 'ODD' Else 'EVEN' end) as Oddity

GO

Using CTE (Recursive logic)

With NumOddity(cnt,dig,num)

as

(

select 1,0,'9075453574568458765456786586578567856785677548964154022106422'

union all

select  cnt+1,convert (int,SUBSTRING(num,LEN(num)+1-cnt,1)), num from   NumOddity where NumOddity.cnt < LEN(num)+1

)

select case (SUM(dig))%2 when 0 then 'EVEN' else 'ODD' End as Oddity from NumOddity

Using Replace function

Based upon the rules

Even  + Even = Even

Even  + Odd  = Odd

Odd   + Odd  = Even

1)  Remove all Even digits from the given number, gives the number containing only Odd digits in the number

2)  Since,   sum of Odd digits is ‘Even’    (If count of the digits is Even)

sum of Odd digits is ‘Odd’     (If count of the digits is Odd)

Ex:

Given number:     9075453574568458765456786586578567856785677548964154022106422

Resulting number, after removing all Evens(0,2,4,6,8) is 9755357557557557575757759151

Count of all digits in the number 9755357557557557575757759151 is, 28, which is Even. So, the sum of the digits in the number becomes Even

select      case len(Replace(Replace(Replace(Replace(Replace('9075453574568458765456786586578567856785677548964154022106422','2',''),'4',''),'6',''),'8',''),'0',''))%2

            when 0 then 'Even' Else 'Odd' End as Oddity