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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- Thread Starter
- 07-03-2013 07:06
- 08-03-2013 04:21
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.