The other tutor has made the error of assuming that the
number with the equal digits must have both its digits
as the same digit as the units digit of the desired number.
This is not a requirement.

Let a = the tens digit
Let b = the units digit

The number = 10a + b

Since the sum of the digits is 14,

a + b = 14, or b = 14 - a

Let c = the digit that is the same in the
answer when 29 is subtracted from the number.
So the answer when 29 is subtracted from the
number is 10c + c or 11c  

10a + b - 29 = 11c

So the system of equations is



This is underdetermined because there are more
unknowns than equations.

We substitute 14 - a for b

10a + (14 - a) - 29 = 11c
10a + 14 - a - 29 = 11c
9a - 15 = 11c

9 is the smallest number in that equation in 
absolute value.  So write 15 and 11 in terms 
of their nearest multiple of 9.  So we write 
15 as 18 - 3 and 11 as 9 + 2

9a - (18 - 3) = (9 + 2)c
9a - 18 + 3 = 9c + 2c

Divide through by 9

a - 2 + 3/9 = c + 2c/9

Get the fractions on one side, 
other terms on the other side:

2c/9 - 3/9 = a - 2 - c 

The right side is an integer, so the left
side is too.  Let that integer be I.

2c/9 - 3/9 = I    a - 2 - c = I; 
    2c - 3 = 9I

2 is the smallest number in that equation in 
absolute value.  So write 3 and 9 in terms 
of their nearest multiple of 2.  So we write 
3 as 2+1 and 9 as 8+1

    2c - 3 = 9I
    2c - (2 + 1) = (8 + 1)I
    2c - 2 - 1 = 8I + I

Divide through by 2

    c - 1 - 1/2 = 4I + I/2

Get the fractions on one side, 
other terms on the other side:

c - 1 - 4I = I/2 + 1/2

The left side is an integer, so the
right side is too.  Let that integer be J.

c - 1 - 4I = J;    I/2 + 1/2 = J
                         I+1 = 2J
                           I = 2J - 1

Substitute for I

c - 1 - 4(2J - 1) = J
   c - 1 - 8J + 4 = J
       c - 8J + 3 = J
                c = 9J - 3

The only value of J that will permit c to be
a digit is J = 1

So c = 9J - 3 = 9(1) - 3 = 6

And since I = 2J - 1,
          I = 2(1) - 1, = 2 - 1 = 1

Now the system of equations becomes:

Since a - 2 - c = I
      a - 2 - 6 = 1
          a - 8 = 1
              a = 9

Since b = 14 - a
      b = 14 - 9 = 5

So the number is 95

Checking:

           95
          -29
           66

Edwin

Let original number’s tens and units digits be T and U, respectively
Let new number’s tens and units digit be 
Then: T + U = 14_____T = 14 - U ------ eq (i)
Also, _____ ------ eq (ii)
 ------ Substituting 14 - U for T in eq (ii)



  

 CANNOT be small digits since that’d make U, or the units digit a 2-digit number, so we start  with the largest digits. 

Substituting 9, 8, and 7 for  into the equation: DOES NOT produce an INTEGER for U that's a multiple of 9. However, 6 did. 


 ------ Substituting 

U = , or 5 

T = 14 - 5 -------- Substituting 5 for U in eq (i)
T, or tens digit = 9

Original number: