The digits of a two-digit number differ by 3. If the digits are interchanged and the resulting number is added to the original number, the sum is 143. What is the original number?

Given :

The digits of a two-digit number differ by 3.

If the digits are interchanged and the resulting number is added to the original number, we get 143.

To do :

We have to find the original number.

Solution :

Let the two-digit number be $10x+y$.

This implies,

$x-y = 3$ or $y-x = 3$

$x = y+3$ or $y = x+3$

The number formed on interchanging the digits is $10y+x$.

Therefore,

$10x+y + 10y+x = 143$

$11x+11y = 143$

If $x = y+3$, then

$11(y+3)+11y = 143$

$11y+33+11y=143$

$22y=143-33$

$22y=110$

$y=\frac{110}{22}$

$y=5$

$x=5+3=8$.

The original number is 85.

If $y=x+3$, then

$11x+11(x+3)=143$

$11x+11x+33=143$

$22x=143-33$

$22x=110$

$x=\frac{110}{22}$

$x=5$

$y=5+3=8$

The original number is 58.

Therefore, the original number can be 58 or 85.

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Updated on: 10-Oct-2022

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