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

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

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.

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

The number.

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

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.

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Answer 2

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.

TO FIND:−

The number.

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 :-

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.

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