Question

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

Answer 1

Whoaaaah....

Answer 2

OK, wait. Maybe not as bad as it first seemed. For the sum of digits to be even, the digits must have an EVEN number of odd digits, or no odd digits.

Answer 3

20 nPr 10 or 20 nCr 10, just plug it in the calculator the reason to this is becasue there are 20 digits but you just want the evens so their are 10 npr is in order and nCr is random.

Answer 4

It seems like a big combinatorics problem??

Answer 5

Hmmm... I'm not sure that is it, because it isn't that you want all even digits? Or 10 even digits. You just need to have an even number of ODD digits??

Answer 6

So we use casework counting? For the numbers

0 odds:...
2odds...
4 odds...

and so on?

Answer 7

@orple8 you have the best questions. This one is very interesting.

Answer 8

Aww... thanks :)

Answer 9

lol i'm sure it's a big help to know how fascinated I am by the question. lol :)

Answer 10

Erin is here, I'll bet she'll know what to do!! :)

Answer 11

wait...why?

Answer 12

I think this question can be approached in this way: Say you pick the first 19 digits. It doesn't matter what they are, or whether their sum is odd or even. But when it comes to choosing the last digit, there are exactly 5 ways to make the sum odd or even (e.g. if the sum of the first 19 digits is even, then we pick an even number).

You have 9 choices for the first digit (as you don't want 0) and 10 choices for the next 18 digits. So the answer is
\[9 \times 10^{18}\times 5 = 450.\]

Answer 13

Okay. That was a failed calculation...

Answer 14

450 seems low for having \(10^{18}\) :P

Answer 15

It was meant to be 45 * 10^18... Ooops :)

Answer 16

Damn. Such an awesomely brilliant explanation, and then she crashed and burned on a power of 10! ;) LOL... (j/k!!) :)

Answer 17

@DebbieG Oh well. It's late. That's my excuse :P

Answer 18

Wow. Just wow. Erin wows again. This was a really hard question though and the explanation was really through. If i could give more than one medal I would. :)

Answer 19

:) I feel your pain! :)

Answer 20

And of course, in general, if you want the number of n-digit numbers there are such that the sum of their digits is even (or odd!), the answer is \(45*10^{n-2}\).

Which is usually not 450.

Answer 21

lol. I have another question that isn't nearly as interesting. I just seem to be really bad at this stuff so...

Answer 22

Just as a final side note, that is assuming that the first digit can't be 0. It's possible that you may be allowed to have numbers that begin with 0.