Question

The sum of the digits of a three-digit number is 11. If the digits are reversed, the new number is 46 more than five times the former number. If the hundreds digit plus twice the tens digit is equal to the units digit, then find the original three-digit number?

Answer 1

Let the hundreds digit be x and the tens digit be ”y” and the unit digit be “z”

∴ The number is 100x + 10y + z

If the digits are reversed the new number is 100z + 10y + x

By the given first condition

x + y + z = 11  ...(1)

By the given second condition

100z + 10y + x = 5(100x + 10y + z) + 46

= 500x + 50y + 5z + 46

x – 500x + 10y – 50y + 100z – 5z = 46

– 499x – 40y + 95z = 46

499x + 40y – 95z = – 46  ...(2)

By the given third condition

x + 2y = z

x + 2y – z = 0  ...(3)

(1) × 95 ⇒    95x + 95y + 95z = 1045   ....(1)
(2) × 1   ⇒  499x + 40y – 95z = – 46     ....(2) 
(1) + (2) ⇒  594x + 135y = 999
                    66x + 15y = 111 (÷ 9)
                    22x + 5y = 37 (÷ 3)          ....(4)
By adding (1) and (3)
                     x + y + z = 11                  ....(1)
                    x + 2y – z = 0                    ....(3)
                       2x + 3y = 11                  .....(5)
(4) × 1   ⇒     22x + 5y = 37                  ....(4)
(5) × 11 ⇒  22x + 33y = 121                 ....(5)
                  (–)   (–)          (–)                    
(4) – (5) ⇒      0 – 28y = – 84

                                y = `84/28` = 3
x = 1, y = 3, z = 7

∴ The number is 137

Subtituting the value of y = 3 in (5)

2x + 3(3) = 11

2x = 11 – 9

2x = 2

x = `2/2` = 1

Subtituting the value of x = 1, y = 3 in (1)

x + y + z = 11

1 + 3 + z = 11

z = 11 – 4

= 7