Question 1199873
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The sum of the digits of a two-digit number is 11. If the digits are reversed, 
the new number is 45 less than the original number. Find the number.
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<pre>
Let "ab" be the decimal presentation of the number, so "b" is the "units" digit and "a" is the "tens" digit.


Then the value of the original number is (10a+b),
while the value of the reversed digit number is (10b+a).


From the problem,

    a + b = 11,                (1)

    10a + b = 10b + a + 45.    (2)


Simplify equation (2) step by step

    10a + b - 10b - a = 45

      9a - 9b = 45

      9(a-b) = 45

        a - b = 45.


Thus we have this system of equations

    a + b = 11,    (3)

    a - b =  5.    (4)


To solve the system, add equations (3) and (4)

    2a = 11 + 5 = 16,  a = 16/2 = 8.


Then from (3)  b = 11 -a = 11 - 8 = 3.


<U>ANSWER</U>.  The number is  83.


<U>CHECK</U>.  The sum of the digits is 8 + 3 = 11;

        the difference of the numbers is  83 - 38 = 45.   ! correct !
</pre>

Solved.