SOLUTION: A certain two-digit number is equivalent to five times the sum of the digits.It is found to be 9 less than the number formed when the digits are interchanged.Find the number.

Let "a" be the tens digit and "b" be the ones digit.


Then the number is 10a+b,  and the first condition gives this equation

    10a + b = 5(a+b),      (1)


which implies

    10a + b = 5a + 5b

     5a     = 4b.          (2)


The solution a = b = 0 is excluded (since "a" can not be zero, as a leading digit).

Then for "a" and "b" being digits in the interval from 1 to 9, equation (1) has one 
and only one solution in integer numbers  a = 4, b = 5.


So, the number is 45, and it is a UNIQUE number, satisfying the first imposed condition.


Thus the second condition is EXCESSIVE and is non-necessary.


It adds nothing to the problem, but we still need check that it is satisfied and 
does not contradict to the first condition.


Fortunately, it is so, and the ANSWER is  45.