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Question

The digit of a two digit number differ by 3. If the digits are interchange and the resulting number is added to the original number we get 143. What can be the original number?

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Solution

Dear student the video cant be added in this section sorry for the inconvinience.the answer is provided with an example for more understandings.check this you will get the answer thank you,

Let us assume, the x is the tenth place digit (for example x=3 in the two digit number 32) and y is the unit place digit of the two-digit number(for example y=2 in the two digit number 32). Also assume x > y

Therefore,

the two-digit number is 10x + y(3cross times10+2=32) and reversed number is 10y + x(2cross times10+3=23);

Given that The digit of a two digit number differ by 3, it can be written as:

x - y = 3.....................(1)

Also given the digits are interchange and the resulting number is added to the original number we get 143, it can be written as:

10x + y + 10y + x = 143

rightwards double arrow11x + 11y = 143

rightwards double arrowx + y = 13 ...............(2)

Adding equation (1) and equation (2) ;

then we get

2x = 16

rightwards double arrowx = 8

then the value of x is put in equation (1),we get;

y = x - 3

= 8 - 3

= 5


Therefore, the two-digit number = 10x + y

= (10 cross times8) + 5

= 85



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