The sum of the digits of a 3-digit number is three times the sum of the digits of a 2-digit number. When the 2-digit number is subtracted from the 3-digit number it gave another 2-digit number of which its first digit is three times the first digit of the first 2-digit number. If the last digit of the 3-digit number equals the last digit of the second 2-digit number, and three times of the sum of the first and second digit of the 3-digit number equals the first digit of the second 2-digit number.
Find the 3-digit number if two times the first digit of the first 2-digit number equals the last digit of the 3-digit number and the sum of the first and second digit of the 3-digit number equals the first digit of the first 2-digit number and also two times the sum of the first and second digit of the 3-digit number equals the last digit of the second 2-digit number. Also the sum of all the digits of the three different number is 27. And the second 2-digit number is greater than the first 2-digit by 66. Finally when the digits of the second 2-digit number are interchanged it is 39 greater than the first 2-digit number.
I don't know please help me
If the 3-digit number is xyz and the two-digit number is ab, then just start listing what you know:
The sum of the digits of a 3-digit number is three times the sum of the digits of a 2-digit number.
x+y+z = 3(a+b)
When the 2-digit number is subtracted from the 3-digit number it gave another 2-digit number of which its first digit is three times the first digit of the first 2-digit number.
(100x+10y+z) - (10a+b) = 10(3a)+c
So now you have two equations for the six variables.
By the time you are done, you should have enough equations to solve for all the variables. See what you come up with from here.
Let's break down the given information and try to solve the problem step by step.
1. The sum of the digits of a 3-digit number is three times the sum of the digits of a 2-digit number.
Let's denote the 3-digit number as XYZ and the 2-digit number as AB. The first clue suggests that X + Y + Z = 3(A + B).
2. When the 2-digit number is subtracted from the 3-digit number, it gives another 2-digit number, of which its first digit is three times the first digit of the first 2-digit number.
If we subtract AB from XYZ, we get a 2-digit number with the first digit three times the first digit of AB. This can be expressed as (X - A) being equal to 3 times the first digit of AB.
3. The last digit of the 3-digit number equals the last digit of the second 2-digit number.
This means that Z is equal to B.
4. Three times the sum of the first and second digit of the 3-digit number equals the first digit of the second 2-digit number.
This can be represented as 3 times (X + Y) being equal to A.
5. Two times the first digit of the first 2-digit number equals the last digit of the 3-digit number.
Let's denote the first digit of AB as A and the last digit of XYZ as Z. This gives us 2A = Z.
6. The sum of the first and second digit of the 3-digit number equals the first digit of the first 2-digit number.
This can be expressed as X + Y = A.
7. Two times the sum of the first and second digit of the 3-digit number equals the last digit of the second 2-digit number.
This can be represented as 2 times (X + Y) equals B.
Now, let's analyze the additional information provided:
- The sum of all the digits of the three different numbers is 27.
- The second 2-digit number is greater than the first 2-digit number by 66.
- When the digits of the second 2-digit number are interchanged, it is 39 greater than the first 2-digit number.
To summarize, we have the following equations:
1. X + Y + Z = 3(A + B)
2. (X - A) = 3A
3. Z = B
4. 3(X + Y) = A
5. 2A = Z
6. X + Y = A
7. 2(X + Y) = B
Now, let's solve these equations:
From equation 4 and 5, we have A = Z = 2A. Therefore, A = Z = 0.
From equation 6, we have X + Y = A = 0. So, X + Y = 0.
From equation 3, we have B = Z = 0.
From equation 7, we have 2(X + Y) = B = 0. Thus, X + Y = 0.
From equation 2, we have X - A = 3A. Since A = 0, we have X = 3A = 3(0) = 0. Therefore, X = 0.
Now, let's substitute these values into equation 1:
0 + 0 + B = 3(A + B)
B = 3(0 + B)
B = 3B
0 = 2B
Since B cannot equal zero, we have a contradiction in our equations.
Therefore, it is not possible to find a valid solution to the given information.
Please double-check the information or provide any missing details to help find a valid solution.