niks18 wrote:

Ditstat wrote:

Hi,

I came up with this method:

In order to sum up the digits to be odd, there are only 3 ways;

1) OOOOO (5^5)

2) EEOOO (4x5^4) as the first digit cannot be zero

3) EEEEO (4*5^4) as the first digit cannot be zero

And I got 13 x 5^4.

What did I do wrong??

Thanks in advance.

I have also approached the problem in the same way. Can any expert clarify as to what am I missing here?

I don't claim to be an expert but I think this may help.

If you want an odd sum, you need an even number of even digits and odd number of odd digits.

Combos of 5 odds (e.g. OOOOO) = 5^5

Total: 5^5

Combos of 3 odds, 2 evens (e.g. EEOOO or OOOEE) = 5!/3!2! = 10

But, if the first number is even, there are only 4 possibilities for that digit (2, 4, 6, 8; 0 will give you a < a 5 digit integer so we can't use it). How many combos of 3 odds, 2 evens will give you an even in the front? E[EOOO] -> 1c1*4c3 = 4 combos with even in front, 10-4=6 combos with odd in front

Numbers with odd digit in front (6 of the 10 combos): 6*(5^5)

Numbers with even digit in front(4 of the 10 combos): 4*(4*5^4)

Total: 6*5^5+4*(4*5^4)

Combos of 1 odd, 4 evens (e.g. EEEEO): = 5!/4! = 5

Numbers with even digit in front (4 of the 5 combos): 4*(4*5^4)

Numbers with odd digit in front (1 of the 5 combos): 1*(5^5)

Sum them all up:

Total numbers with odd digit in front: 1*5^5+6*5^5+1*5^5 = 8*5^5 = 25,000

Total numbers with even digit in front: 4*(4*5^4)+4*(4*5^4)=8*4*5^4=20,000

Sum: 45,000 = 45*10^3 = D