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OpenStudy (anonymous):
How many 20-digit numbers are such that the sum of their digits is even?
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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
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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
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OpenStudy (anonymous):
however we cannot choose "0" as first digit , so only 9 choices for first digit
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