Math
In a 3 digit number the hundreds digit is one half of tens digit while the ones digit is more than the tens digit. If the sum of 3 digit is 11, find the numbers.
The tens digit must be even, since the hundreds is 1/2 of it.
with the ones digit greater, that leaves
128 or 245
I suspect a typo.
To find the three-digit numbers that satisfy these conditions, we can break down the problem into steps.
Step 1: Set up the problem
Let's assume the hundreds digit is 'x', the tens digit is 'y', and the ones digit is 'z'.
Step 2: Translate the given information into equations
From the given information, we have two equations:
1. The hundreds digit is one-half of the tens digit: x = (1/2)y
2. The ones digit is more than the tens digit: z > y
Step 3: Determine the sum of the digits
We know that the sum of the digits in a three-digit number is 11. So, we can set up another equation: x + y + z = 11.
Step 4: Solve the system of equations
Now we will solve the system of equations to find the values of x, y, and z.
From equation 1, we can rewrite it as y = 2x.
Substituting this value into equation 3, we get:
x + (2x) + z = 11
3x + z = 11
Step 5: Test different values to find the solution
We need to test different values of x and z that satisfy the equation 3x + z = 11. We also need to ensure that z is greater than y.
By substituting different values, we find:
For x = 1, z = 8 (satisfies both equations)
For x = 2, z = 5 (satisfies both equations)
For x = 3, there is no value of z that satisfies the condition z > y.
Therefore, we have found two valid three-digit numbers that satisfy the given conditions:
1. The first number is 185 (hundreds digit = 1, tens digit = 8, ones digit = 5).
2. The second number is 265 (hundreds digit = 2, tens digit = 6, ones digit = 5).
So, the numbers are 185 and 265.