The Sum of the Other Rearrangement of Digits

Suppose that the number is $1000a+100b+10c+d$ .

There are $24$ different rearrangements of the digits ( including itself ). Notice that there are $6$ arrangements when the thousand-digit is $a$ , $6$ arrangements when the thousand-digit is $b$ , $6$ arrangements when the thousand-digit is $c$ , $6$ arrangements when the thousand-digits is $d$ . Same for the hundreds-digit, tens-digit, and units-digit.

Thus, the sum of the $24$ rearrangements is $6000(a+b+c+d)+600(a+b+c+d)+60(a+b+c+d)+6(a+b+c+d)=6666(a+b+c+d)$ . This means that the sum of all the rearrangements of the digits of a four-digit number (including itself) is divisible by $6666$ .

We are told that the $f(x)=62339$ . This is the sum of the rearrangements that does not include the $x$ . So, if $x$ is added to $62339$ , the number is divisible by $6666$ (as discussed above). Since we are finding the smallest possible value for $x$ , we need to find the smallest positive value that, when added to $62339$ , is divisible by $6666$ .

The smallest number divisible by $6666$ that is larger than $62339$ is $66660$ . Thus, the smallest possible value of $x$ is $x=66660-62339=4321$ .

We can check that $4321$ indeed works. The sum of all 24 arrangements of $4321$ is $6666(4+3+2+1)=66660$ . Now, this minus $4321$ is $66660-4321=62339$ .