In a three-digit number, when the tens and the hundreds digit are inte

1.	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» Let the three digits numbers be 100a + 10b + c. 100b + 10a + c = 3(100a + 10b + c) + 54 … (1) 100a + 106 + c + 198 = 100c + 106 + a … (2) (b – a) = 2(b – c) … (3) (1) ⇒ 100b + 10a + c = 300a + 30b + 3c + 54 ⇒ 290a – 70b + 2c = -54 (2) ⇒ 99a – 99c = -198 ⇒ a – c = -2 ⇒ a = c – 2 (3) ⇒ a + b – 2c = 0 ⇒ a + b = 2c ⇒ b = 2c – c + 2 ⇒ b = c + 2 Substituting a, b in (1) 290(c – 2) – 70 (c + 2) + 2c = -54 290c – 580 – 70c – 140 + 2c = -54 222c = 666 ⇒ c = 3 a = 1, 6 = 5 ∴ The number is 153.

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»

Let the three digits numbers be 100a + 10b + c.

100b + 10a + c = 3(100a + 10b + c) + 54 … (1)

100a + 106 + c + 198 = 100c + 106 + a … (2)

(b – a) = 2(b – c) … (3)

(1) ⇒ 100b + 10a + c = 300a + 30b + 3c + 54

⇒ 290a – 70b + 2c = -54

(2) ⇒ 99a – 99c = -198 ⇒ a – c = -2

⇒ a = c – 2

(3) ⇒ a + b – 2c = 0 ⇒ a + b = 2c

⇒ b = 2c – c + 2

⇒ b = c + 2

Substituting a, b in (1)

290(c – 2) – 70 (c + 2) + 2c = -54

290c – 580 – 70c – 140 + 2c = -54

222c = 666 ⇒ c = 3

a = 1, 6 = 5

∴ The number is 153.