# How many numbers are there in range 1 to 1000 which contains digits 2 and 3 and divisible 2 and 3?

How many numbers are there in range 1 to 1000 which contains digits 2 and 3 and divisible 2 and 3?
I know the answer to find count of numbers in range 1 to 1000 which are divisble by 2 and 3. But the above question also asks that those numbers must include digits 2 and 3.
Please genaralise your answer beacuse I want to know the answers when there are more or different digits. Like find count of numbers in range 1 to 1000000 which contains digit 5,2,7,8 and divisible by 5,2,7,8.

• You're looking for a multiple of 6 whose decimal representation has a 2 and a 3. There are no such one or two digit numbers. For three digit numbers, you must have $2+3+x$ divisible by 3, where $x$ is the third digit. So $x$ is 1, 4, or 7. But you also need the number to be even. Hence... ? – symplectomorphic Jul 4 '16 at 8:11
• As I see, there's probably no simple way to generalise. Counting is probably the simplest way to do it – Aditya De Saha Jul 4 '16 at 8:26
• I think the problem was specific and wanted you to rely on the fact that the third digit must be 1, 4 or 7 (else it's not divisible by 3) and the number must end in 4 (else it's not divisible by 2). This is certainly not extendable to general cases. – fleablood Jul 6 '16 at 5:59
• Here is the source of the problem: cs.stackexchange.com/q/60267/755 – D.W. Jul 10 '16 at 16:58

$3$-digit numbers containing $[2,3]$ and divisible by $[2,3]$:

• $132$
• $234$
• $312$
• $324$
• $342$
• $372$
• $432$
• $732$

$6$-digit numbers containing $[2,5,7,8]$ and divisible by $[2,5,7,8]$:

• $175280$
• $258720$
• $357280$
• $387520$
• $538720$
• $567280$
• $572880$
• $580720$
• $728560$
• $735280$
• $752080$
• $758240$
• $785120$
• $807520$
• $857920$
• $875280$
• $877520$
• $958720$
• I am not looking for specifically this case. Can you please generalise your answer? If you have to find how many numbers are there in range 1 to 1000000000 having digits a,b,c and divisible by digits a,b,c. where 1<=a,b,c<=9 and a!=b!=c – Mike Jul 4 '16 at 8:25
• @Mike: That depends on the values of $[a,b,c]$. – barak manos Jul 4 '16 at 8:28

A general exact answer seems difficult, but here's a way to get a good estimate that readily generalises. Divisibility by $2$ constrains the last digit; divisibility by $3$ doesn't constrain any digits and can be treated as roughly independent of the other constraints. Thus, if we treat the problem probabilistically, we have the events $D_i$ of containing digit $i$ and $E_i$ of divisibility by $i$, we're looking for $\textsf{Pr}(D_2\cap D_3\cap E_2\cap E_3)$, and we can treat $E_3$ as independent of the three other events and $E_2$ as dependent with $D_2$ and $D_3$ only through the last digit. Then

\begin{align} \def\pr#1{\textsf{Pr}\left(#1\right)} \pr{D_2\cap D_3\cap E_2\cap E_3} &=\pr{D_2\cap D_3\cap E_2}\pr{E_3} \\ &= \frac13\left(\pr{D_2\cap E_2}+\pr{D_3\cap E_2}-\pr{(D_2\cup D_3)\cap E_2}\right) \\ &= \frac13\left(\frac12\left(1-\frac45\left(\frac9{10}\right)^{n-1}\right)+\frac12\left(1-\left(\frac9{10}\right)^{n-1}\right)-\frac12\left(1-\frac45\left(\frac8{10}\right)^{n-1}\right)\right) \\ &= \frac16\left(1-\frac{2\cdot9^n-8^n}{10^n}\right)\;, \end{align}

where $n$ is the number of digits, so the expected count of such numbers would be

$$\frac16\left(10^n-2\cdot9^n+8^n\right)\;.$$

In your case, with $n=3$, this is

$$\frac16\left(10^3-2\cdot9^3+8^3\right)=9\;,$$

in reasonable agreement with the fact that barak manos counted $8$.

For the case of $5$, $2$, $7$ and $8$, divisibility by $5$ and $2$ simply means that the number ends in a $0$; this is equivalent to asking how many numbers with $n-1$ digits contain $5$, $2$, $7$ and $8$ and are divisible by $7$ and $4$. Again divisibility by $7$ is independent since $7$ is coprime to the base $10$, but now the calculation is slightly more involved since divisibility by $4$ constrains the last $2$ digits and you need to perform inclusion-exclusion on all $4$ digit values that should be present. Still, the ingredients are all in the calculation above.


1. $\ds{c}$ is an even number and $3 \mid \pars{a + b + c}$.
2. If $\ds{\color{#f00}{c} \not= \color{#f00}{2}}$, then $\ds{a + b = 5}$. It leaves us with $\ds{c = 4}$ because $\ds{3 | \pars{5 + 0}}$,$\ds{3|\pars{5 + 6}}$ and $\ds{3|\pars{5 + 8}}$ are false.
3. If $\ds{\color{#f00}{c} = \color{#f00}{4}}$, it obviously leaves us with $\ds{\ul{two}}$ numbers: $\ds{\color{#f00}{234}}$ and $\ds{\color{#f00}{324}}$.
4. If $\ds{\color{#f00}{c} = \color{#f00}{2}}$, $\ds{a + b + 2 = 3p}$ where $\ds{p}$ is an integer $\ds{~\pars{\mbox{for the time being,}\ 1 \leq p \leq 6}~}$. Because $\ds{a = 3}$ or $\ds{b = 3}$, one of them is equal to $\ds{3p - 2 -3 = 3p - 5}$. Note that $\ds{3p - 5 \not= 3}$. \begin{align} &\mbox{But}\quad 0 \leq 3p - 5 \leq 9\quad\imp\quad {5 \over 3} \leq p \leq {14 \over 3}\quad\imp\quad p \in \braces{2,3,4} \\ &\imp\quad 3p - 5 \in \braces{1,4,7}\ \mbox{which yields}\ 3\times 2 = 6\ \mbox{numbers}:\ \left\lbrace\begin{array}{l} \ds{\color{#f00}{132}} \\ \ds{\color{#f00}{312}} \\ \ds{\color{#f00}{342}} \\ \ds{\color{#f00}{432}} \\ \ds{\color{#f00}{372}} \\ \ds{\color{#f00}{732}} \end{array}\right. \end{align}
There are $\ds{\color{#f00}{8}}$ numbers which satisfy the OP conditions. Namely, $$\begin{array}{|c|}\hline\mbox{}\\ \ds{\quad% \color{#f00}{132},\color{#f00}{234},\color{#f00}{312},\color{#f00}{324},\color{#f00}{342},\color{#f00}{372},\color{#f00}{432},\color{#f00}{732} \quad} \\ \mbox{}\\ \hline \end{array}$$