Question 1179510
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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.
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<pre>
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 <U>ANSWER</U> is  45.
</pre>

Solved.