The sum of a two digit number and the number formed by interchanging its digits is 110. If 20 is subtracted from the original number, the new number is 4 more than 4 times the sum of the digits in the original number. Find the original number.

Hello!
The original number may be written as bar(ab), where a and b are one-digit natural numbers, 0lt=blt=9, 1lt=alt=9. The value of the number bar(ab) is obviously b + 10a.
When we interchange its digits, the resulting number is bar(ba) = a + 10b. The sum is (b + 10a) + (a + 10b) = 11a + 11b = 110 (this is given). So
a + b = 10.
The second condition gives that
(b + 10a) - 20 = 4(a + b) + 4, or 6a - 3b = 24, or 2a - b = 8.
Thus we have a linear system for a and b. If we add up these two equations, we obtain 3a = 18, or a = 6. From the first equation b = 10 - a = 4. The initial restrictions for a and b are also satisfied.
The answer: the original number is 64.