Let the tens and units digits of her age be T and U

Then we get: 

10U + T + 17 = 20T + 2U ------ Cross-mutiplying

8U = 19T - 17



I decided to solve for U (units digit) as opposed to T (tens digit) because in most cases, the tens digit is smaller, and I can arrive at the answer much quicker 

Substituting 1 for T results in 0.25 for U, which obviously is NOT an integer

Substituting 2 for T results in 2.625 for U, which obviously is NOT an integer

However, substituting 3 for T results in a value of 5 for U, which obviously is an INTEGER, and so we get 

I also used Excel and no other digit for T, from 4 - 9, produces an INTEGER-DIGIT for U. 

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!
 


Let t and u be the tens and units digits of her age.  Then


10t+u = her age

10u+t = her age with the digits reversed

(10u+t+17)/2 = her age, with the digits reversed, and 17 added, and the whole thing divided by 2


The problem says that expression is equal to her age:







This is a single equation with two variables -- an example of a Diophantine equation.  There are an infinite number of solutions; however, there is only one solution that satisfies the conditions of the problem -- that t and u are single digit positive integers.


To solve a Diophantine equation, solve the equation for one variable in terms of the other:









To find single digit positive integer solution(s) for that equation, perform the division on the right to write the expression as a whole number and a remainder:





In that equation, u and (2t-2) are integers, so (3t-1)/8) has to be an integer.


The only single digit positive integer t for which that expression is an integer is t=3.  So







ANSWER: Emma's age is 35


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


NOTE: You can solve part or all of this problem by trial and error, knowing that there are very few options for the tens and units digit of her age.  Other tutors have provided responses to your post in which they do that.


There is of course nothing wrong with solving the problem that way; if getting an answer as quickly as possible is important, then that is probably the best way to solve the problem.


However, knowing how to solve linear Diophantine equations is a useful skill to have; so I thought it useful to show this formal method for solving the problem.


 

Answer by Edwin McCravy(20059)   (Show Source): You can put this solution on YOUR website!
 
Their answers are correct, but they all essentially used trial and error
to solve the Diophantine equation in integers.  Here is how to do it
by actually solving for t and u, without using trial and error.  

[Warning: it is much longer, but more practical in general integer problems.]

In fact, your teacher may not allow trial and error, since one of the main
uses of algebra is to avoid trial and error.

(1)        8b + 17 = 19a

The smallest coefficient of either letter in absolute value is 8.
We write every number in terms of the nearest multiple of 8.

       8b + (16+1) = (16+3)a

       8b + 16 + 1 = 16a + 3a

Divide through by 8

       b + 2 + 1/8 = 2a + 3a/8  

Isolate the fractions:

       b + 2 - 2a = 3a/8 - 1/8

The left side is an integer, so the right side is also. 
Let that integer be P.  So we have:

(2)    b + 2 - 2a = P

and

       3a/8 - 1/8 = P

Clearing of fractions:

(3)        3a - 1 = 8P

The smallest coefficient of either letter in absolute value is 3.
We write every number in terms of the nearest multiple of 3.

           3a - 1 = (6+2)P

           3a - 1 = 6P + 2P

Divide through by 3

          a - 1/3 = 2P + 2P/3  

Isolate the fractions:

           a - 2P = 2P/3 + 1/3

The left side is an integer, so the right side is also. 
Let that integer be Q.  So we have:

(4)        a - 2P = Q

and

       2P/3 + 1/3 = Q

Clear of fractions:

(5)        2P + 1 = 3Q

---------

The smallest coefficient of either letter in absolute value is 2.
We write every number in terms of the nearest multiple of 2.

           2P + 1 = (2+1)Q

           2P + 1 = 2Q + Q

Divide through by 2

          P + 1/2 = Q + Q/2  

Isolate the fractions:

            P - Q = Q/2 - 1/2

The left side is an integer, so the right side is also. 
Let that integer be R.  So we have:

(5)         P - Q = R

and

        Q/2 - 1/2 = R

Clear of fractions:

            Q - 1 = 2R

(6)             Q = 2R + 1

We finally have solved for a letter with no fraction terms, so
we can now substitute back:

Use (6) to substitute for Q in 5

(5)         P - Q = R
     P - (2R + 1) = R
       P - 2R - 1 = R
                P = 3R+1

(4)        a - 2P = Q
      a - 2(3R+1) = 2R + 1
       a - 6R - 2 = 2R + 1
(6)             a = 8R+3

Since a is a digit,

      0  a <= 9
      0 <= 8R+3 <= 9
     -3 <= 8R <= 6
   -3/8 <= R <= 6/8

Since R is an integer, R = 0

Substituting in (6)

(6)   a = 8R+3
      a = 8(0)+3
      a = 0+3
(7)   a = 3

Substitute in 

(1)   8b + 17 = 19a  
      8b + 17 = 19(3)
      8b + 17 = 57
           8b = 40
            b = 5

So Emma's age = 35

Now I will agree that their trial and error methods were much easier and
much shorter. However, in general, Diophantine equations, trial and error
would involve trying many integers.  In those cases this method would be the
only practical method. 

Edwin
 

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