The digits of a four-digit positive integer add up to 14. The sum of the two middle digits is nine, and the thousands digit minus the units digit is one. If the integer is divisible by 11, what is the integer?
+26367 The digits of a four-digit positive integer add up to 14.
The sum of the two middle digits is nine, and the thousands digit minus the units digit is one.
If the integer is divisible by 11,
what is the integer?
\(\begin{array}{|lrcll|} \hline \text{A four-digit positive integer}:& abcd \\ \hline \text{The digits add up to $14$}: & a+b+c+d=14 \\ \text{The sum of the two middle digits is nine}: & b+c=9 \\ \text{The thousands digit minus the units digit is one}: & a-d=1 \\ \text{Divisibility by Eleven: if the difference of the alternating sum is $0$}: & a+c-(b+d)=0 \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline a+b+c+d &=& 14 \quad | \quad b+c=9 \\ a+9+d &=& 14 \quad | \quad a-d=1\quad \text{or} \quad a=1+d \\ 1+d+9+d &=& 14 \\ 2d+10&=& 14 \\ 2d &=& 4 \\ \mathbf{d} &=& \mathbf{2} \\ \hline a &=& 1+d \\ a &=& 1+2 \\ \mathbf{a} &=& \mathbf{3} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline a+c-(b+d) &=& 0 \\ a+c-b-d &=& 0 \\ a-d+c-b &=& 0 \quad | \quad a-d = 1 \\ 1+c-b &=& 0 \quad | \quad b+c=9\quad \text{or} \quad c=9-b \\ 1+9-b-b &=& 0 \\ 2b &=& 10 \\ \mathbf{b} &=& \mathbf{5} \\ \hline c &=& 9-b \\ c &=& 9-5 \\ \mathbf{c} &=& \mathbf{4} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \text{The four-digit positive integer $abcd$ is}:& \mathbf{3542} \\ & 3542 : 11 = 322 \\ \hline \end{array} \)
