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.

Question

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 just need all the equation I will solve it my self

I do connexus · Accepted Answer

uhh how many questions is that

Kenny · Answer

I have tried honestly am not getting it I just need all the equation I will go through the pain solving them