Any two-digit number can be written as 10x + 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...
Any two-digit number can be written as 10x + 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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