$x$ is a positive integer such that its digits can only be $3$,$4$,$5$, $6$. $x$ contains at least one copy of each of these four digits. The sum of the digits of $x$ is $900$ and the sum of the digits of $2x$ is also $900$. How many digits are there in the maximum value of $x$?
$3\times2=6\\4\times2=8\\5\times2=10\\6\times2=12$
Tens digit can be maximum $1$.
If in $x$ there was ___ $\rightarrow$ upon multiplication in $2x$, ___
$3\rightarrow6, \space\space3\rightarrow7\\4 \rightarrow 8, \space\space4\rightarrow9 \\5 \rightarrow 0, \space\space 5 \rightarrow1\\6\rightarrow2, \space\space 6\rightarrow3$
First, we have to establish the basics:
3 multiplied 2 gives 6.
6 multiplied 2 gives 12, which has a sum of 3.
Note that $6+3=3+6$ ---> sum does not change.
4 multiplied 2 gives 8.
5 multiplied 2 gives 10, which has a sum of 1.
Note that $4+5=8+1$ ---> sum does not change.
Observation: Since the sum of digits are the same for $x$ and $2x$, we can conclude that there are equal number of 3s and 6s, and equal number of 4s and 5s, such that the sum remains the same.
Assume we have $m$ number of 3, and $n$ number of 4. Then we have $3m+6m+4n+5n=900$. $m+n=100$.
Now we want to find $2m+2n$, which is the number of digits, which is $200$. Do note that the smallest $x$ has also $200$ digits.