A three digit number is equal to 17 times the sum of its digits. If 198 is added to the number , the digits are interchanged . The addition of first and third digit is 1 less than middle digit. Find the number. [5 marks]

Answer:

Let the ones place digit be z

tens place digit be y

hundredth place digit be x

then , the original number = 100x + 10y + z

Interchanged number = 100z+ 10y + x

According to the 1st condition,

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

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

100x – 17x + 10y- 17y + z- 17z=0

83x -7y -16z =0 ——————A

According to 2nd condition,

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

100x +10y + z-100z – 10y – x = -198

99x – 99z = -198

Dividing by 99 both sides,

x – z = -2 i.e x +2 = z —————B

According to 3rd condition,

x +z = y – 1

x + x +2 = y – 1 (From— B)

2x + 2+ 1 =y

2x + 3 = y ————————-C

Substitute values of Z and Y from equation B and C in equation A,

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

Put x = 1 in equation B

x + 2 = z

1 + 2 = z

z = 3

Put x= 1 in equation C

2x + 3 = y

2(1) + 3 =y

y = 5

We have x= 1; y =5 ; z=3

three digit number = 100x + 10y + z

= 100 (1) + 10 (5) + 3

= 100 + 50 + 3

= 153