Question

In a three-digit number, when the tens and the hundreds digit are interchanged the new number is 54 more than three times the original number. If 198 is added to the number, the digits are reversed. The tens digit exceeds the hundreds digit by twice as that of the tens digit exceeds the unit digit. Find the original number

Answer 1

Let the hundred digit be x

the tens digit be y and the unit digit be z

∴ The number is 100x + 10y + z

By the given first condition

100y + 10x + z = 54 + 3 (100x + 10y + z)

100y + 10x + z = 54 + 300x + 30y + 3z

– 290x + 70y – 2z = 54 .......(÷ – 2)

145x – 35y + z = – 27  ...(1)

Again by the second condition

198 + 100x + 10y + z = 100z + 10y + x

99x – 99z = – 198 .......(÷ 99)

x – z = – 2  ...(2)

Again by the third condition

y – x = 2(y – z)

y – x = 2y – 2z

– x – y + 2z = 0

x + y – 2z = 0   ...(3)

(1) × (2) ⇒ 290x – 70y + 2z = – 54  ...(1)
(3) × (1) ⇒       x +    y – 2z  =  0     ....(3)
(1) + (3) ⇒ 291x – 69y = – 54
                   97x – 23y = – 18   ....(4)
(3) × 1        x + y – 2z = 0       ....(3)
(2) × 2      2x + 0 – 2z = – 4    ....(2)
                  (–)  (+)  (+)   (+)   
(3) – (2) ⇒   – x + y = 4          ....(5)
(4) × 1 ⇒      97x – 23y = – 18 ....(4)
(5) × 3 ⇒   – 23x + 23y = 92   ....(5)
(4) × (5) ⇒              74x = 74
                             x = `74/74` = 1

Substitute the value of x = 1 in .....(2)

1 – z = – 2

3 = z

∴ z = 3

Substitute the value of x = 1 and z = 3 in .......(3)

1 – y – 6 = 0

y – 5 = 0

y = 5

∴ The number is 153