Question 482070
There is no "standard formula."


You have to use some algebra by giving some digit a name. Suppose the units digit is u. Then the tens digit is u-3. The original number is 6 more than 4 times the sum of the digits means:


*[tex \LARGE 10(u-3) + u = 6 + 4((u-3) + u)]


(I simply converted the statement into an equation, with the "original number" being 10(u-3) + u).


Solving for u,


*[tex \LARGE 10(u-3) + u = 6 + 4(2u - 3)]


*[tex \LARGE 11u - 30 = 8u - 6] (simply expand both sides)


*[tex \LARGE 3u = 24]


*[tex \LARGE u = 8]


Hence, the units digit is 8 and the tens digit is 5. We can check since 58 is 6 more than the 4 times the sum of the digits (13).