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.
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
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