Question

the difference between a 2-digit number are the number formed
by reversing its digit 45. If the sum of the digits of the Original number is 13 , then find the number.​

Answer 1

EXPLANATION.

Let the digit at tens place be = x

Let the digit at unit place be = y

original number = 10x + y

reversing number = 10y + x

The difference between a 2 digit

number and the number formed by

reversing it's digit = 45.

=> ( 10x + y) - ( 10y + x) = 45

=> 10x + y - 10y - x = 45

=> 9x - 9y = 45

=> x - y = 5 .......(1)

The sum of digit of original number = 13.

=> x + y = 13 ........(2)

From equation (1) and (2) we get,

=> 2x = 18

=> x = 9

Put the value of x = 9 in equation (2) we get,

=> 9 + y = 13

=> y = 4

Therefore,

original number = 10x + y = 10(9) + 4 = 94.

Answer 2

Given ,

• The difference between a two digit number and the number formed by reversing its digit 45

• The sum of the digits of the original number is 13

We know that ,

If the digits of two digits number are x (ones digit) and y (tens digit) , then the required number is

$\boxed{ \tt{10y + x}}$

Thus ,

Original number = 10y + x

New number = 10x + y

Since ,

Original number - New number = 45

Thus ,

10y + x - (10x + y) = 45

9y - 9x= 45

y - x = 5 --- (I)

Now , the sum of the digits of the original number is 13

Thus ,

x + y = 13 --- (II)

Adding eq (I) and eq (II) , we get

y - x + x + y = 5 + 13

2y = 18

y = 18/2

y = 9

Put y = 9 in eq (I) , we get

9 - x = 5

x = 4

Hence , the original number is 94

_________________ Keep Smiling ☺️