sum of 3 number is equal 39 , the second number is half of third number ,and the first number less than third number by 1,the result must be 6 degit ?

what "result" must be 6 digits?

x+y+z = 39
y = z/2
x = z-1

x=15
y=8
z=16

To find the three numbers that satisfy the given conditions, we can set up a system of equations and solve them simultaneously.

Let's denote the three numbers as follows:
- First number: x
- Second number: y
- Third number: z

According to the given conditions:
1. The sum of the three numbers is equal to 39:
x + y + z = 39

2. The second number is half of the third number:
y = z/2

3. The first number is less than the third number by 1:
x = z - 1

Now, let's solve this system of equations to find the values of x, y, and z.

Substitute the value of y from equation 2 into equation 1:
x + (z/2) + z = 39
Multiplying the equation by 2 to eliminate fractions:
2x + z + 2z = 78
2x + 3z = 78 ----(Equation 4)

Substitute the value of x from equation 3 into equation 4:
2(z - 1) + 3z = 78
2z - 2 + 3z = 78
5z - 2 = 78
5z = 80
z = 80/5
z = 16

Substitute the value of z into equation 3 to find x:
x = z - 1 = 16 - 1 = 15

Substitute the value of z into equation 2 to find y:
y = z/2 = 16/2 = 8

So, the three numbers are:
x = 15
y = 8
z = 16

Now, let's check if the summation of the three numbers is equal to 39:
15 + 8 + 16 = 39

Therefore, the three numbers that satisfy the given conditions are 15, 8, and 16. Their sum is indeed equal to 39.

Since each of these numbers is in the range of 0 to 99, the result is indeed a 6-digit number, when represented without leading zeros.