Question

(ii) A three-digit number is equal to 17 times the sum of the digits. If the digits are
reversed the new number is 198 more than the original number. The sum of the
extreme digits is 1 less than the middle digit. Find the original number
​

Answer 1

let the numbers be xyz

three digit number obtained by them: 100x + 10y + z

according to first information in the question,

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

100x + 10y + z = 17x + 17y + 17z

83x - 7y - 16z = 0.......(1)

by reversing the order, number will be zyx

three digit number obtained by them: 100z + 10y + x

according to second information in the question,

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

99x - 99z = -198

99(x - z ) = -198

x-z = -2

z = x + 2...............(2)

according to third information in the question,

x + z = y - 1

x + x + 2 = y - 1

2x +2 = y - 1

2x - y = -3

y = 2x + 3.........(3)

putting value of y and z in eq...(1)

83x - 7y - 16z = 0

83x - 7(2x+3) - 16(x+2) = 0

83x - 14x - 21 - 16x - 32 = 0

83x - 30x - 53 = 0

53x = 53

x=1

so, z = x +2

z = 1+2 = 3

and

y = 2x + 3

y = 2(1) + 3

y = 2 + 3

y = 5

Hence, the three digit number is 153.