mathmath333
  • mathmath333
How many different 7 digit members are their sum of whose digits are odd ?
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{ How many different 7 digit members are their sum }\hspace{.33em}\\~\\ & \normalsize \text{ of whose digits are odd ?}\hspace{.33em}\\~\\ & a.)\ 45\times 10^{5} \hspace{.33em}\\~\\ & b.)\ 24\times 10^{5} \hspace{.33em}\\~\\ & c.)\ 224320 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{ none of these } \hspace{.33em}\\~\\ \end{align}}\)
ParthKohli
  • ParthKohli
odd + odd = even even + even = even odd + even = odd we need an odd number of odd digits
ParthKohli
  • ParthKohli
only 1 odd, 3 odd, 5 odd are allowed

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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & 5\ odd +2\ even \hspace{.33em}\\~\\ & 3\ odd +4\ even \hspace{.33em}\\~\\ & 1\ odd +6\ even \hspace{.33em}\\~\\ \end{align}}\)
ParthKohli
  • ParthKohli
first consider the cases where the first digit is taken by an odd number... (i) 1 odd number O _ _ _ _ _ _ the only odd number is in the first digit. 1, 3, 5, 7, 9 are the possible choices. the rest of the digits are even. so 0, 2, 4, 6, 8 are the choices for those. overall, \(5^7\) numbers in this category (ii) 3 odd numbers O _ _ _ _ _ _ first we choose the two remaining places in \(\binom{6}{2}\) ways. now we have \(5\) choices for each of those three places for odd numbers and \(5\) choices for all other places too. so the count is \(\binom{6}{2}\times 5^7 \) for this category. (iii) 5 odd numbers O _ _ _ _ _ _ first choose the four remaining places in \(\binom{6}{4}\) ways, then multiply by \(5^7\)
ParthKohli
  • ParthKohli
now consider the cases where an odd number does NOT occupy the first place E _ _ _ _ _ _ the possible choices for the first digit are 2, 4, 6, 8 and the rest of the even digits are 0, 2, 4, 6, 8. so then (i) one odd number E _ _ _ _ _ _ choose the place for the odd-number in \(\binom{6}{1}\) ways first. for the first place, 4 choices. all other six places - five choices. so \(\binom{6}{1} \times 4 \times 5^6\) ways. (ii) three odd numbers E _ _ _ _ _ _ \(\binom{6}{3}\) ways to choose the place. now the count is \(\binom{6}{3}\times 4 \times 5^6\) (iii) five odd numbers E _ _ _ _ _ _ \(\binom{6}{5}\) ways to choose the place. now the count is \(\binom{6}{5}\times 4 \times 5^6\)
ParthKohli
  • ParthKohli
The answer is approximately A.
ParthKohli
  • ParthKohli
\[44.21875 \times 10^5\]that's more like what I get.
ParthKohli
  • ParthKohli
Maybe I've not counted something, because the answer is exactly A.
mathmath333
  • mathmath333
is this true there are 9 * 10^6 seven digit numbers and half of them have sum of digits odd
mathmate
  • mathmate
@parthkohli If we count 7 odd numbers which gives 5^7=78125. 4421875+78125=4500000, bingo!
ParthKohli
  • ParthKohli
that's the stupidest mistake i could have made
ParthKohli
  • ParthKohli
we can also prove that there's a bijection from the set of odd sum-of-digits to even sum-of-digits to prove that the cardinality of the two sets is the same.
mathmate
  • mathmate
Then we need to accept 7 digit numbers with leading zeroes!

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