Determine a 2-Digit Number Satisfying Two Conditions

Linear algebra problems and solutions

Problem 649

A 2-digit number has two properties: The digits sum to 11, and if the number is written with digits reversed, and subtracted from the original number, the result is 45.

Find the number.

 
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Solution.

The key to this problem is noticing that our 2-digit number can be written as $10A + B$, where $A, B$ are the 10s and 1s digit respectively.

The digits summing to 11 then yields the equality $A + B = 11$.

The number with digits reversed is $10B + A$, and so the second property yields the equation $10A + B – (10B + A) = 45$. Simplifying, we have the system of equations
\begin{align*}
A+B &= 11\\
9A-9B &= 45
\end{align*}


To solve this system, we create the augmented matrix and use elementary row operations to put it into reduced row-echelon form:
\begin{align*}
\left[\begin{array}{rr|r} 1 & 1 & 11 \\ 9 & -9 & 45 \end{array} \right] \xrightarrow{ R_2 – 9 R_1 } \left[\begin{array}{rr|r} 1 & 1 & 11 \\ 0 & -18 & -54 \end{array} \right]\\[6pt] \xrightarrow{ \frac{-1}{18} R_2 } \left[\begin{array}{rr|r} 1 & 1 & 11 \\ 0 & 1 & 3 \end{array} \right] \xrightarrow{R_1 – R_2} \left[\begin{array}{rr|r} 1 & 0 & 8 \\ 0 & 1 & 3 \end{array} \right].
\end{align*}
With this final augmented matrix, we can read off the solution $A = 8$ and $B = 3$.

Thus our number is $10A + B = 83$.


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