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

Solution :

Let the required three digit number be abc , a, b and c are all non zero .

Now , the sum of the first two digits is 4 .

So,

a + b = 4 . . . (1)

Now , the sum of the numbers which are by using each digit of the number once is 1998.

That means :

The sum of all possibilities distinct permutations of the number abc without repeatation is 1998 .

=> abc + acb + bac + bca + cab + cba = 1998

=>( 100 a + 10 b + c ) + ( 100 a + 10 c + b ) + ( 100b + 10 a + c ) + ( 100 b + 10 c + a ) + ( 100c + 10 a + b ) + ( 100c + 10b + a ) = 1998

=> ( 100 a + 100a + 10 a + a + 10a + a ) + ( 100 b + 100b+ 10 b + b + 10b + b ) + ( 100 c + 100c + 10 c + c + 10c + c ) = 1998

=> 222 a + 222b + 222c = 1998

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

=> a + b + c = 9 .

Now , a + b = 4 .

So , c = 5 .

Thus , the number becomes :

=> ab5

where a + b = 4 , 0 < a , b ≤ 9 .

Now , we need to find the possible values of a and b such that ab5 has 8 factors .

The possible values of (a,b ) => ( 1,3 ) , ( 2, 2 ) , ( 3, 1 ) .

Thus , the possible numbers become :

=> 135 ; 225 ; 315 .

All of these three numbers have 4 pairs of factors or 8 factors .

Answer :

The possible numbers are 135 , 225 and 315 respectively .

_____________________________________

Answer 2

$\Large\underline{\underline{\sf{ \color{magenta}{GIVEN:-} }}}$

• 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.
• The number has 8 factors.

$\Large\underline{\underline{\sf{ \color{magenta}{TO\ FIND:-} }}}$

• The number.

$\Large\underline{\underline{\sf{ \color{magenta}{SOLUTION:-} }}}$

Let the three digit number be abc.

A/q,

a + b = 4                . . .(i)

Again, the sum of the numbers which are by using each digit of the number once is 1998.

According to the above given statement, we get :-

$\sf{abc + acb + bac + bca + cab + cba = 1998}$

Putting the places' value and adding further :-

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

→ 200a + 200b + 200c + a + 10a + a + b + 10b + b + c + 10c + c = 1998

→ 200a + 22a + 200b + 22b + 200c + 22c = 1998

→ 222a + 222b + 222c = 1998

→ a + b + c = 9

Putting the value of a + b from equation (i) :-

4 + c = 9

→ c = 5

Possible values for a and b (c = 5):

a = 1, b = 3

a = 2, b = 2

a = 3, b = 1

The required number could be :-

135, 225 or 315

A/q, the required number must have 8 factors.

◘ 135 is the only number which has 8 factors.

So, a = 1 and b = 3.

Again, putting the places' value and adding in abc :-

abc = 100a + 10b + c

→ abc = 100(1) + 10(3) + 5

→ abc = 100 + 30 + 5

→ abc = 135

Hence, the answer is 135.