When you reverse the digits in a certain two-digit number you decrease its value by 9. Find the number if the sums of its digits is 14

Question

I seriously just CAN'T get this one at all!

Hope you can help!

Answer 1

Let the number be :  T U

where T is the tens digit and U is the ones(units) digit.

For example, 23 would have T=2 and U=3.

The value of the number is  T*10 + U

Now reverse those digits:   U T

This new number has a value:  U*10 + T

which is 9 less than the original number/value.

So that gives the equation:

    original number = reversed number + 9

   T*10 + U  =  U*10 + T  + 9

 

The other requirement is that the sum of the digits is 14, so:

  T + U = 14

 

Now we have two equations in two unknowns:

   9T = 9U + 9   -->  T = U + 1

   T + U = 14

Substituting for T in the second equation gives:

  U+1 + U = 14

  2U = 13

    U = 6.5   Oooops...

 

It seems there is no solution to the problem as given...

We can check the possibilities of digits that add to 14:

 

 T  U    U T    TU - UT

 9  5    5 9      36

 8  6    6 8      18

 7  7    7 7       0

 6  8    8 6     -18

 5  9    9 5     -36

 

This may explain why you can't get it: it can't be gotten ;-)

 

Answer 2

Actually, a 2-digit number can be represented by a formula because Decimal is a positional number system.

 

Let x=first digit

     y=second digit

 

Putting them together:

  The two-digit number is  (10x+y).

 

The number in reverse is  (10y+x).  This represents a decrease of 9, so

    (10x+y) - 9 = (10y+x)     [eq1]

 

The sum of the digits is 14 means  x+y=14   [eq2]

 

  9x -  9y =     9             [from eq1]

  9x + 9y = 126             [9*eq2]

-------------------     [elimination; add equations]

 18x        =  135

    x         =  7.5    [oops !!  this is not a whole number]

 

So --

 list the digits that add to 14

   5   9

   6   8

   7   7

   8   6

   9   5

 

Is 59-95=9 ? no

Is 68-86=9 ? no

Is 77-77=9 ? no

Is 86-68=9 ? no

Is 95-59=9 ? no

 

That explains your difficulty !  There is no answer.

 

Back to the drawing board.  PLZ verify that you have the problem correctly stated.