Question

How many different 7 digit members are their sum
of whose digits are odd ?

Answer 1

\(\large \color{black}{\begin{align}
& \normalsize \text{ How many different 7 digit members are their sum }\hspace{.33em}\\~\\
& \normalsize \text{ of whose digits are odd ?}\hspace{.33em}\\~\\
& a.)\ 45\times 10^{5} \hspace{.33em}\\~\\
& b.)\ 24\times 10^{5} \hspace{.33em}\\~\\
& c.)\ 224320 \hspace{.33em}\\~\\
& d.)\ \normalsize \text{ none of these } \hspace{.33em}\\~\\
\end{align}}\)

Answer 2

odd + odd = even
even + even = even
odd + even = odd

we need an odd number of odd digits

Answer 3

only 1 odd, 3 odd, 5 odd are allowed

Answer 4

\(\large \color{black}{\begin{align}
& 5\ odd +2\ even \hspace{.33em}\\~\\
& 3\ odd +4\ even \hspace{.33em}\\~\\
& 1\ odd +6\ even \hspace{.33em}\\~\\
\end{align}}\)

Answer 5

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 \(5^7\)

Answer 6

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 \(\binom{6}{5}\times 4 \times 5^6\)

Answer 7

The answer is approximately A.

Answer 8

\[44.21875 \times 10^5\]that's more like what I get.

Answer 9

Maybe I've not counted something, because the answer is exactly A.

Answer 10

is this true there are 9 * 10^6 seven digit numbers
and half of them have sum of digits odd

Answer 11

@parthkohli
If we count 7 odd numbers which gives 5^7=78125.
4421875+78125=4500000, bingo!

Answer 12

that's the stupidest mistake i could have made

Answer 13

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.

Answer 14

Then we need to accept 7 digit numbers with leading zeroes!