Note that "$1,3,5,7,9 = 5!\times 10^5$" is nonsense as written. I don't know what the left hand side is supposed to be, but $5!\times 10^5$ is not equal to $9$, and is not equal to the sequence of the five odd numbers $1$, $3$, $5$, $7$, and $9$.
And the count is completely off, as many have mentioned.
One possibility is to use inclusion-exclusion: there are $10^{10}$ possible phone numbers; how many have no $9$s in it? $9^{10}$; same for numbers with no $7$s, no $5$s, no $3$'s, or no $1$s. So one first approximation is
$$10^{10} - \binom{5}{1}9^{10}.$$
However, we are double counting the numbers that have no $9$s and no $7$s, the ones that have no $9$s and no $5$s, etc. There are $8^{10}$ of each of the $\binom{5}{2}=10$ pairs of odd digits. So we adjust by adding those back in,
$$10^{10}-\binom{5}{1}9^{10} + \binom{5}{2}8^{10}.$$
But now we are off with those that are missing three of the odd numbers; we took them out three times with $\binom{5}{1}9^{10}$, but we then added them back three times with $\binom{5}{2}8^{10}$; we need to take them out again. So we get
$$10^{10}-\binom{5}{1}9^{10} + \binom{5}{2}8^{10} - \binom{5}{3}7^{10}.$$
But now we are off with the numbers that are missing four of the odd numbers: we took them out four times with $\binom{5}{1}9^{10}$; we added them back in six (that is $\binom{4}{2}$) times with $\binom{5}{2}8^{10}$; then we took them out four times (that is, $\binom{4}{3}$) with $\binom{5}{3}7^{10}$. So we need to add them back in, with $\binom{5}{4}6^{10}$, so we get
$$10^{10} - \binom{5}{1}9^{10} + \binom{5}{2}8^{10} - \binom{5}{3}7^{10} + \binom{5}{4}6^{10}.$$
But now, what about those numbers that fail to contain all of the odd digits? We counted them five times in the $\binom{5}{1}9^{10}$ summand, so we took them out five times; then we added them back in $\binom{5}{2}=10$ times in the summand $\binom{5}{2}8^{10}$; then we took them out $10$ times in $\binom{5}{3}7^{10}$; and then we added them back in five times with $\binom{5}{4}6^{10}$. In total, we've took them out zero times; we need them out, so we need to subtract $\binom{5}{5}5^{10}$. So, finally, we get
$$10^{10} - \binom{5}{1}9^{10} + \binom{5}{2}8^{10} - \binom{5}{3}7^{10} + \binom{5}{4}6^{10} - \binom{5}{5}5^{10}$$
telephone numbers which contain each of $1$, $3$, $5$, $7$, and $9$ at least once.
Let $s$ be the number of $5$-digit numbers that are multiples of $3$.
Let $m$ be the number of $5$-digit numbers that are multiples of $3$ and don't contain $6$.
Finding $s$ is easy, it is $90000/3=30000$
Finding $m$ is also easy, there are exactly three multiples of $3$ that don't contain $6$ for every possible "start" (because each residue appears three times among the digits different from $6$)
. So if $ \overline{abcd}$ is a four digit number there are $3$ digits $x$ so that $\overline{abcdx}$ is a multiple of $3$ not containing $6$. so $m=(8\times9\times9\times9)\times 3=17496$
Hence the answer is $30000-17496=12504$
Best Answer
There are $9\cdot10^4$ five-digit numbers, since there are $9$ choices for the first digit, and $10$ choices for the next $4$ digits.
Of these numbers, $8\cdot 9^4$ do not contain $4$ as a digit. Therefore the number of $5$-digit numbers with at least one $4$ as a digit is equal to $$ 9\cdot 10^4-8\cdot 9^4=37512$$