The sum of the digit of a certain two-digits number 12. When you revenue it's digits you increase the number by 54. Find the number

Question

A math word problem that is very hard and needs to be answered 

Answer 1

Look at the way our counting system works:

From right to left we have 1's, 10's, 100's, 1000's, etc..

21 = 2(10) + 1

235 = 2(100) + 3(10) + 5

and so forth....  

 

Let the number be xy

 

The value of the original number is:  10x + y

If we reverse the digits to yx the value is 10y + x

 

We know:  x + y = 12    as we are told this in the problem.

 

Reversing the digits increases the value by 54 so:

10y + x = 10x + y + 54

 

Moving all the variables to the left in the 2nd equation we have:

-10x - y + 10y + x = 54

-9x + 9y = 54

9(-x + y) = 54

-x + y = 54/9

-x + y = 6

 

So we have two equations with x and y:

 

 x + y = 12

-x + y = 6       add the equations to eliminate x

----------------

     2y = 18

      y = 9

 

If y = 9 and x + y = 12 then x = 3

 

The original number is 39

The reversed number is 93

 

Answer 2

You can solve this by algebra, provided you make this correction to the statement of the problem:

...when you REVERSE ITS DIGITS. [It's is only used to mean IT IS].

 

Here is how to begin:   Let the units digit be u, let the tens digit be 5.

 

10u + t = 10t + u +54; also t+u=12.