Reverse Digits

Let the two digit integer be expressed as $10x+y$ where $x$ and $y$ are its two digits. Reversing the integer yields $10y+x$ , which must be 20% greater. Therefore, $\frac{10y+x}{10x+y}=\frac{6}{5}$ . By multiplying by the denominators, $50y+5x=60x+6y$ , and hence $44y=55x$ . This simplifies further to $4y=5x$ , so $y=\frac{5}{4}x$ . $x$ and $y$ are both integers between 1 and 10, so the only way to satisfy this is $x=4$ and $y=5$ . The original integer must hence be $10\times 4 + 5 = \boxed{45}$ .

Another solution is -if we don't know the steps then see for only reversible 2 digit number -whichs 20% is less than 10:; Like 12-21,,23-32,,34-43,,45-54,,56-65..... So find 20% of each number and see it is reversed or not This is easiest way to find answer So answer is 45

20% more = 120% of the original number. New:Old = 120:100 = 6:5. So the old number (integer) has to be a multiple of 5, yet not ending in zero, as the other wouldn't be a 2 digit number. One could do it algebraically, but there's only nine numbers to check brute force style. Less knowing the outliers dont have a chance.

Let two digit integer is $\overline{ab}$ = 10a + b

So, 10b + a = 10a + b + 20%(10a + b)

10b + a = 10a + b + $\frac{1}{5}$ (10a + b)

50b + 5a = 50a + 5b + 10a + b

55a = 44b

5a = 4b

So, a = 4 and b = 5

The two integer is 45

I quite enjoyed that puzzle.

10a + b (x1.2) = 10b + a

12a + 1.2b = 10b + a

11a = 8.8b

It then is a simple step to ensure that 8.8b is an integer with b <10

Still a nice little puzzle

10T + U = 1.20 ( 10U + T) U/T = 4/5 45 is the two digit number