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 will solve the equation My self
Just help me interpret please have been trying and trying
Let's break down the given information and solve each part 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 assume the 3-digit number is ABC, where A, B, and C represent the hundreds, tens, and units digits, respectively. Similarly, let's assume the 2-digit number is XY, where X and Y represent the tens and units digits, respectively.
The sum of the digits of ABC is A + B + C, and the sum of the digits of XY is X + Y. According to the given information, we have the equation:
A + B + C = 3(X + Y) ----(1)
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.
When XY is subtracted from ABC, we get the 2-digit number PQR. Here, P, Q, and R represent the tens and units digits of the resulting number.
To represent this information as an equation, we have:
ABC - XY = PQR
(100A + 10B + C) - (10X + Y) = 10P + Q + R
100A + 10B + C - 10X - Y = 10P + Q + R ----(2)
Likewise, we have the additional information that the first digit of PQR is three times the first digit of XY, giving us another equation:
10P + Q + R = 3X ----(3)
3. The last digit of the 3-digit number equals the last digit of the second 2-digit number.
The units digit of ABC is equal to the units digit of PQR, which can be expressed as:
C = R ----(4)
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.
To represent this, we have:
3(A + B) = P ----(5)
5. The sum of all the digits of the three different numbers is 27.
From this, we can deduce:
(A + B + C) + (X + Y) + (P + Q + R) = 27
(A + B + C) + (X + Y) + (3X) = 27
(A + B + C) + (4X + Y) = 27 ----(6)
6. The second 2-digit number is greater than the first 2-digit number by 66.
This gives the equation:
XY + 66 = PQR ----(7)
7. When the digits of the second 2-digit number are interchanged, it is 39 greater than the first 2-digit number.
This can be represented as:
YX + 39 = XY ----(8)
Solving these equations simultaneously will lead us to the values of all the variables and ultimately the 3-digit number.