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.
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
"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
Let's start by assigning variables to the digits of the three-digit number. Let's say the hundreds digit is "x", the tens digit is "y", and the units digit is "z".
According to the problem, the sum of the digits is 11:
x + y + z = 11 ----(1)
We are also given that if we reverse the order of the digits, the number is decreased by 396. This means the original number minus 396 is equal to 100z + 10y + x:
100z + 10y + x = 100x + 10y + z - 396 ----(2)
We are also given that the tens digit is half of the hundreds digit:
y = (1/2)x ----(3)
Now we can solve these three equations simultaneously to find the values of x, y, and z.
First, let's substitute the value of y from equation (3) into equations (1) and (2):
x + (1/2)x + z = 11 ----(4)
100z + 10(1/2)x + x = 100x + 10(1/2)x + z - 396 ----(5)
Simplifying equation (4):
(3/2)x + z = 11 ----(6)
Simplifying equation (5):
100z + 5x + x = 100x + 5x + z - 396
100z + 6x = 105x + z - 396
99z = 99x - 6x + 396
Subtracting 99x from both sides:
99z - 99x = -6x + 396
99(z - x) = -6x + 396
(z - x) = (-6x + 396) / 99
(z - x) = -x/11 + 4
Since z and x are digits, the left side of the equation must be an integer. Thus, -x/11 must also be an integer. The only possible values of x that work are -11, 0, or 11. However, x must be a positive digit, so we can disregard -11.
If x = 0, from equation (6) we get:
(3/2)(0) + z = 11
z = 11
But z cannot be 11 because it was given that z is a digit.
Therefore, we can conclude that x must be 11.
Substituting x = 11 into equation (6) we get:
(3/2)(11) + z = 11
(33/2) + z = 11
33 + 2z = 22
2z = -11
z = -11/2
But z must be a digit, so this solution is not valid.
Therefore, there is no valid solution for this problem.
To solve this problem, we will use algebraic equations and a systematic approach.
Let's start by assigning variables to the unknown digits of the three-digit number. We'll call the hundreds digit "x," the tens digit "y," and the units digit "z." Therefore, the original three-digit number can be expressed as 100x + 10y + z.
Now, we can translate the given information into equations:
1) The sum of the digits of a three-digit number is 11: x + y + z = 11.
2) If the order of the digits is reversed, the number is decreased by 396: (100z + 10y + x) = (100x + 10y + z) - 396.
3) The tens digit is one half of the hundreds digit: y = 0.5x.
Now, let's solve these equations systematically:
Step 1: Substitute y = 0.5x from equation 3 into equation 1:
x + 0.5x + z = 11.
Simplifying the equation, we get 1.5x + z = 11.
Step 2: Rewrite equation 2 using the variables:
100z + 10y + x = 100x + 10y + z - 396.
Simplifying the equation, we get 99z = 99x - 396.
Step 3: Divide equation 2 by 99 to eliminate the variables:
z = x - 4.
Now, let's substitute this value of z into equation 1:
1.5x + (x - 4) = 11.
Simplifying the equation, we get 2.5x = 15.
Dividing both sides by 2.5, we get x = 6.
Using this value of x, we can find y:
y = 0.5x = 0.5 * 6 = 3.
Finally, we can substitute the values of x and y into equation 3 to find z:
z = x - 4 = 6 - 4 = 2.
So, the hundreds digit is 6, the tens digit is 3, and the units digit is 2. Therefore, the three-digit number is 632.