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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

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2 answers
  1. uhh how many questions is that

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  2. I have tried honestly am not getting it

    I just need all the equation I will go through the pain solving them

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