Any two-digit number can be written as 10*x + y, *where x is the tens digit and y is the unit (ones) digit. This is because y has the place value of 1 and x has the place value of 10. The value of x can be any integer between 1 and 9, and the value of y can be any integer between 0 and 9.

According to the problem, the tens digit exceeds the unit digit by 3. This can be written, in term of x and y, as x = y + 3.

Also, the sum of the digits is 1/7 of the number itself. This means that

`x+y = 1/7(10x + y)`

Now we have two equations with two variables, that can be solved for x and y:

x = y + 3 and `x+y = 1/7 (10x + y)`

Before solving, the second equation can be simplified by multiplying both right and left side by 7. It then becomes

7(x+y) = 10x + y

7x + 7y = 10x + y

-3x + 6y = 0

The two equations are now x = y + 3 and -3x + 6y = 0. This system can be solved by substitution. Substitute x = y + 3 from the first equation into the second:

-3(y + 3) + 6y = 0

-3y -9 + 6y = 0

3y = 9

y = 3

If y = 3, then x = y + 3 = 3 + 3 = 6

So the **number is then 63**. Notice that both conditions are satisfied: the tens digit exceeds the unit digit by 3, and the sum of digits (9) is 1/7 of the number: 9 is 1/7 of 63.

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