The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting numbers is added to the original number, we get 143. Find both the numbers.

SOLUTION

Define the digits

Let the tens digit be \(x\) and the units digit be \(y\).
Then the original number is
\(10x + y\)
The number obtained by interchanging the digits is
\(10y + x\)

Use the sum condition

The sum of the original number and the interchanged number is \(143\). \( (10x + y) + (10y + x) = 143\)
⇒ \( 10x + x) + (10y + y) = 143\)
⇒ \( 11x+ 11y = 143\)
⇒ \( 11(x + y) = 143\)
⇒ \( x + y = 13\) --- (I)

Use the digit difference condition

The digits differ by 3: \(|x − y| = 3\)
There are two possible cases:
(i)\(x − y = 3\)
(ii)\(y − x = 3\) (or \( x − y = −3\))

Solve Case (i)

We have, \(x + y = 13\) and \(x − y = 3\)
Adding the two equations, we get
\(2x = 16\) ⇒ \(x = 8\)
Substituting into \(x + y = 13\), we get
\(8 + y = 13\) ⇒ \(y = 5\)
So the original number is \(10x + y = 10(8) + 5 = 85\)

Solve Case (ii)

We have, \(x + y = 13\) and \(x − y = −3\)
Adding the two equations, we get
\(2x = 10\) ⇒ \(x = 5\)
Substituting into \(x + y = 13\), we get
\(5 + y = 13\) ⇒ \(y = 8\)
So the original number is \(10x + y = 10(5) + 8 = 58\)

🏆

Final Answer : The numbers are 58 and 85.

Concept Note

A two-digit number is of the form \(10x + y\), where \(x\) is the tens digit and \(y\) is the units digit. Reversing the digits give a new number \(10y + x\). Use this knowledge and the conditions given to form appropriate equations and solve for \(x\) and \(y\).