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    If a certain number of two digit be divided by the sum of the digits, the quotient is 2 and the remainder 8.Three times the given number is 2 greater than the number expressed by the digit in reversed order. What is the number?
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    Some hints for you:

    Let the two digits (from left to right) be a and b.

    Then the number formed is given by 10a+b

    Since this number when divided the sum of the digits yield a quotient of 2 and a remainder of 8,

    then we have (10a+b)/(a+b)= 2 + 8/(a+b) =======> 10a+b = 2(a+b) +8 ----------(1)

    Based on the second set of information provided in the problem, formulate a separate equation relating a and b similar to (1), before proceeding to solve things simultaneously.

    Hope this helps. Peace.
 
 
 
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