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Mathematics 14 Online
OpenStudy (anonymous):

How many 20-digit numbers are such that the sum of their digits is even?

OpenStudy (tmak11):

oh dear...this may take a while

OpenStudy (anonymous):

even plus even causes even and odd plus odd causes even

OpenStudy (anonymous):

we need even number of odd numbers in order for the number to be positive

OpenStudy (tmak11):

so all odd numbers for all 20 digits would be evan aka 20!

OpenStudy (anonymous):

you can make the sum even by fixing the last digit

OpenStudy (tmak11):

true

OpenStudy (anonymous):

So total 20 digit numbers with even parity would be : 9*10^18*5

OpenStudy (anonymous):

9 ways to choose first digit 10 ways to choose 2nd through 19th dgitis 5 ways to choose last digit

OpenStudy (tmak11):

wouldn't you just use that thing were you take 20 and multiply it by 19 then mult that by 18 and so in

OpenStudy (anonymous):

there are 10 digits in base10 system so there are 10 digits to choose from for each position

OpenStudy (anonymous):

however we cannot choose "0" as first digit , so only 9 choices for first digit

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