dtan5457
  • dtan5457
If O represents the number of integers between 10000 and 100000 all whose digits are odd, and E represents all the number of integers between 10000 and 100000 all of whose digits are even, what is the value of O-E?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
All numbers within the range have 5 digits. For any one digit, you have 5 choices (1,3,5,7, or 9), so \(O=5^5\). You can determine \(E\) similarly, but keep in mind that the first digit can't be \(0\).
dtan5457
  • dtan5457
Oh, I get it now. I thought it meant the amount of odd numbers between 10000 and 100000, which would be a lot more
dtan5457
  • dtan5457
So for E, even numbers, you would have 4 choices (2,4,6,8)

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anonymous
  • anonymous
Right, if we're only counting odd numbers we only need to check the last digit. The first digit of every number in \(E\) has 4 choices, yes (2,4,6, or 8), but every digit after that can also be a \(0\), so those have 5 choices.
dtan5457
  • dtan5457
And since the other digits can be 0, that should be 5^4
dtan5457
  • dtan5457
so in total O=5^5 E=4x5x5x5x5
anonymous
  • anonymous
Correct
dtan5457
  • dtan5457
and from there it's just basic subtraction
dtan5457
  • dtan5457
thanks man
anonymous
  • anonymous
No problem. You can avoid actually computing the powers, though, and instead make a slight manipulation: \[5^5-4\times5^4=5^4(5-4)=5^4\]

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