How many 20-digit numbers are such that the sum of their digits is even?
oh dear...this may take a while
even plus even causes even and odd plus odd causes even
we need even number of odd numbers in order for the number to be positive
so all odd numbers for all 20 digits would be evan aka 20!
you can make the sum even by fixing the last digit
true
So total 20 digit numbers with even parity would be : 9*10^18*5
9 ways to choose first digit 10 ways to choose 2nd through 19th dgitis 5 ways to choose last digit
wouldn't you just use that thing were you take 20 and multiply it by 19 then mult that by 18 and so in
there are 10 digits in base10 system so there are 10 digits to choose from for each position
however we cannot choose "0" as first digit , so only 9 choices for first digit
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