Question

brain time Q7.There is a three-digit number. The second digit is four times as big as the third digit, while the first digit is three less than the second digit. What is the number?​

Answer 1

Answer:

$-300, 141, 582$

Step-by-step explanation:

Let: the first digit of the number be $x$

      the second digit of the number be $y$

      the third digit of the number be $z$

The question states that:

$i)$ The second digit is four times as big as the third digit

   ⇒ $y = 4z$                  --$(1)$

$ii)$ The first digit is three less than the second digit

    ⇒ $x = y-3 = 4z - 3$         --$(2)$

Since, both equation are dependent on $z$  

Lets take conditions based on $z$

$Case$ $I:$   $z = 0$

           ⇒ $4z = y = 0$

           ⇒ $x = 4z-3 = -3$

           ∴ The number becomes $-300$

$Case$ $II:$    $z = 1$

             ⇒ $4 z = y = 4$

             ⇒ $x = 4z - 3 = 1$

             ∴ The number becomes $141$

$Case$ $III:$    $z = 2$

               ⇒ $4z = y = 8$

               ⇒ $x = 4z - 3 = 5$

               ∴ The number becomes $582$

$Case$ $IV:$    $z = 3$

              ⇒ $4z = y = 12$

                  Any digit of a number cannot be a two-digit number

              ∴ This case cannot be possible

              ⇒ $z<3$

$Case$ $V:$    $z = -1$

                 Any digit other than the first-digit cannot be a negative number

            ⇒ $z>0$    

            ∴ This case is not possible.

⇒ We are left with three numbers $-300, 141, 582$

∴ The three-digit numbers whose second digit is four times as big as the third digit and the first digit is three less than the second digit are:  $-300, 141, 582$