+0  
 
0
96
1
avatar

A, B and C each represent a single digit with no two digits being the same. Find the values of A, B and C so that the following is true: AB + C=38 BC+A=29

 Aug 4, 2020
 #1
avatar+25597 
+1

A, B and C each represent a single digit with no two digits being the same.
Find the values of A, B and C so that the following is true: \(AB + C=38\) and \(BC+A=29\)

 

My attempt:

\(\begin{array}{|rcll|} \hline \mathbf{AB + C} &=& \mathbf{38} \\ 10A+B+C &=& 38 \qquad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{BC+A} &=& \mathbf{29} \\ 10B+C+A &=& 29 \qquad (2) \\ \hline \end{array} \\ \begin{array}{|lrcll|} \hline (1)-(2): & 10A+B+C -(10B+C+A) &=& 38-29 \\ & 10A+B+C - 10B -C -A &=& 38-29 \\ & 10A+B - 10B -A &=& 9 \\ & 9A-9B &=& 9 \quad | \quad :9 \\ & A-B &=& 1 \\ & \mathbf{B} &=& \mathbf{A-1} \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline (1): & \mathbf{10A+B+C} &=& \mathbf{38} \quad | \quad \mathbf{B=A-1} \\ & 10A+A-1+C &=& 38 \\ & 11A-1+C &=& 38 \\ & 11A +C &=& 39 \\ & \mathbf{ 11A } &=& \mathbf{39-C} \\ \hline \end{array}\)

 

\(\begin{array}{|c|c|c|} \hline C & 39-C & A = \dfrac{39-C}{11} \\ \hline 0 & 39 & \text{not integer} \\ \hline 1 & 38 & \text{not integer} \\ \hline 2 & 37 & \text{not integer} \\ \hline 3 & 36 & \text{not integer} \\ \hline 4 & 35 & \text{not integer} \\ \hline 5 & 34 & \text{not integer} \\ \hline \color{red}6 & 33 & \mathbf{A =3} \\ \hline 7 & 32 & \text{not integer} \\ \hline 8 & 31 & \text{not integer} \\ \hline 9 & 30 & \text{not integer} \\ \hline \end{array}\), so \(\mathbf{C=6}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{B} &=& \mathbf{A-1} \quad | \quad \mathbf{A =3} \\ B &=& 3-1 \\ \mathbf{B} &=& \mathbf{2} \\ \hline \end{array}\)

 

check:

\(\begin{array}{|rcll|} \hline 32+6 &=& 38 \qquad ( AB+C = 38) \\ 26+3 &=& 29 \qquad ( BC+A = 29) \\ \hline \end{array}\)

 

laugh

 Aug 4, 2020
edited by heureka  Aug 4, 2020

60 Online Users

avatar
avatar
avatar