Hello everybody,
we've been working on this problem in class: "How many 6 digit natural numbers there are with even sum of digits?".
I thought this was fairly simple and straightforward problem:
So the number of numbers would be 9*10*10*10*10*5.
But the examiner told me I need to count the number of even and odd digits and approach the problem completely differently without explaining what's wrong with mine. I tried to find some flaws in it, but I can't. Why is it more complicated than it seems? Thank you!
we've been working on this problem in class: "How many 6 digit natural numbers there are with even sum of digits?".
I thought this was fairly simple and straightforward problem:
- The first digit can be 1-9 (not 0, we are considering decimals); so 9 options
- Second to fifth can be any; so 10 options
- Sum of any 5 natural number (first 5 digits) is either even or odd natural number, so in order the "even stays even" and "odd becomes even", we need to add even last digit for even sum and odd last digit to odd sum; so exactly half possible numbers in both cases - 5 options
So the number of numbers would be 9*10*10*10*10*5.
But the examiner told me I need to count the number of even and odd digits and approach the problem completely differently without explaining what's wrong with mine. I tried to find some flaws in it, but I can't. Why is it more complicated than it seems? Thank you!
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