Question

The sum of the first two digits of a three digit number is 4. The sum of the numbers
that is formed by using each digit of the number once is 1998. If the number has 8
factors, then what is the number?

Answer 1

Let the number is $\overline{abc}.$ Then

$a+b=4, a\not=0$

$$100a+10b+c+100a+b+10c+$$

$$+100b+10a+c+100b+a+10c+$$

$$+100c+10a+b+100c+a+10b=1998$$

$$222(a+b+c)=1998$$

$$a+b+c=9$$

$$c=9-4=5$$

The possible numbers are

$$135, 225, 315, 405$$

The 8 factors of 135 are: $1,3,5,9,15,27,45,135$

The 9 factors of 225 are: $1,3,5,9,15,25,45, 75,225$

The 12 factors of 315 are: $1,3,5, 7,9,15,21, 35, 45, 63,105, 315$

The 10 factors of 405 are: $1,3,5, 9,15,27, 45, 81,1305, 405$

Therefore the number is $135.$