Question

A three digit number is equal to 17 times the sum of its digits if 198 is added to the number the digits are reversed the sum of the first and the third digit is 1 less than the middle digit find the number

Answer 1

Let the hundredth, tenth and ones digit of the given number of x, y, and z

Then the number will be

100x + 10y + z

The number is 17 times the sum of digits

⇒ 100x + 10y + z = 17 (x + y + z) = 17x + 17y + 17

⇒ 100x – 17x + 10y – 17y + z – 17z = 0

⇒ 83x – 7y – 16z = 0 ... (1)

Also if 198 is added to the number the digits sets reversed.

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

⇒ 100x – x + 10y – 10y + z – 100z + 198 = 0

⇒ 99x – 99z + 198 = 0

⇒ x – z + 2 = 0

⇒ z = x + 2 ... (2)

The the sum of the extreme digits is less than the middle digit by unity.

⇒ x + z = y – 1

= x – y + z + 1 = 0 ... (3)

Putting the value of z from (2) in (1) we get

83x – 7y – 16 (x + 2) = 0

⇒ 83x – 7y – 16x – 32 = 0

⇒ 67x – 7y = 32 ... (4)

Putting the value of z from (2) in (3) we get

x – y + x + 2 + 1 = 0

⇒ 2x – y + 3 = 0

⇒ y = 2x + 3 ... (5)

Putting the value of y from (5) in (4) we get

67x – 7 (2x + 3) = 32

67x– 14x – 21 = 32

53x = 32 + 21 = 53

⇒ x = 1 ... (6)

Putting the value of x from (6) in (8) we get

y = 2 + 3 = 5 ... (7)

Putting the value of x and y from (6) and (7) in (3) we get

1 – 5 + z + 1 = 0

⇒ z = 3

Hence the required number is 153.