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**Hint:**if there is a two digit number it means there are two places, ones and tens. For solving questions let assume the digit at once place be x and tense place be y. Then apply the condition of the question to solve further.

**Complete step-by-step answer:**

Here in the given question there is a two digit number whose sum is 13.

Now, let the digit at once place be x and tense place be y.

Then the sum of the digit is x+y=13…… let it equation (1), as in the question it is given that the sum of digit numbers is 13.

Tens place=y

Once place= x

Therefore the two digit number forms by these two digits is

Number= 10(y)+1(x) as we know that for forming a two digit number we multiply the digit at tens place by 10 and then add it with the digit at once place by multiplying 1 at once place.

Therefore the required number is 10y+x.

Now according to question we interchange the numbers,

I.e. tense place become= x and once place became=y

The new number formed is, Number= 10(x) +1(y).

Therefore new number=10x+y.

Now in the question, it says that if the number is subtracted from the one obtained by interchanging the digits, the result is 45.

i.e. 10x+y-(10y+x) =45

$ \Rightarrow 9x - 9y = 45$ Take 9 common and divide it to the right hand side.

$ \Rightarrow x - y = 5$……….let it equation (2)

On adding (1) and (2) we get,

2x=18

$\therefore x = 9$

Now put the value of x in equation (1) we get,

9+y=13

$\therefore y = 4$

Therefore the two digit number$ = 10y + x = 10 \times 4 + 9 = 49$.

**Hence, the required number is 49.**

**Note:**Whenever we face such a type of question the key concept for solving the question is first assume the digits at once place be x and tense place be y and then form the numbers by these digits according to the question. And then apply the statement of question to proceed a d then we will find the value of that digit. Now put the value of that digit to get the number.

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