The sum of the digits of a three-digit number is 11. If the order of the digits is reversed, the number is decreased by 396. The tens digit is one half of the hundreds digit. Find the number.

Answer 1

a = hundreds digit

b = tens digit

c = ones

Original number is:

100 a + 10 b + c

The sum of the digits of a three-digit number is 11 means:

a + b + c = 11

The tens digit is one half of the hundreds digit means:

b = a / 2

a + b + c = 11

a + a / 2 + c = 11

3 a / 2 + c = 11

Subtract 3 a / 2 to both sides.

c = 11 - 3 a / 2

When order of the digits is reversed, reverse number is:

100 c + 10 b + a

The number is decreased by 396 means:

original number - reverse number = 396

100 a + 10 b + c - ( 100 c + 10 b + a ) = 396

100 a + 10 b + c - 100 c - 10 b - a = 396

99 a - 99 c = 396

Replace c by 11 - 3 a / 2 in this equation.

99 a - 99 ∙ (11 - 3 a / 2 ) = 396

99 a -1089 + 297 a / 2 = 396

Add 1089 to both sides.

99 a + 297 a / 2 = 1485

198 a / 2 + 297 a / 2 = 1485

495 a / 2 = 1485

Multiply both sides by 2

495 a = 2970

a = 2970 / 495

a = 6

c = 11 - 3 a / 2 = 11 - 3 ∙ 6 / 2 = 11 - 18 / 2 = 11 - 9

c = 2

a + b + c = 11

6 + b + 2 = 11

b + 8 = 11

b = 11 - 8

b = 3

The number is:

100 a + 10 b + c = 100 ∙ 6 + 10 ∙ 3 + 2 = 600 + 30 + 2 = 632

Check result.

Reverse number is:

100 c + 10 b + a = 236

original number - reverse number = 632 - 236 = 396

The tens digit is one half of the hundreds digit:

b = a / 2

3 = 6 / 2

Answer 2

"The tens digit is one half of the hundreds digit"
---> let the tens digit be x, then the hundreds digit is 2x
let the unit digit by y
"The sum of the digits is 11" ----- 2x + x + y = 11, or
3x + y = 11 **

original number: 100(2x) + 10x + y or 210x + y
number reversed: 100y + 10x + 2x or 100y + 12x

210x+y - (100y + 12x) = 396
198x - 99y = 396 , divide by 99
2x - y = 4 ***

add ** and ***
5x = 15
x = 3, then from ** , y = 2

the original number was 632

check:
632 - 236 = 396 , my answer is correct