If we divide a 2-digit positive integer by the sum of its digits, we get the quotient and remainder of 4 and 3, respectively.
If we divide the same 2-digit positive by the product of its digits, we get quotient and remainder of 3 and 5, respectively.
What is the 2-digit integer?
+26367 If we divide a 2-digit positive integer by the sum of its digits,
we get the quotient and remainder of 4 and 3, respectively.
If we divide the same 2-digit positive by the product of its digits,
we get quotient and remainder of 3 and 5, respectively.
What is the 2-digit integer?
The 2-digit integer \(ab\) is \(10a + b\)
\(\begin{array}{|lrcll|} \hline & \begin{array}{rcll} \text{We divide a 2-digit positive integer}\\ \text{by the sum of its digits} \\ \end{array} \\ \hline (1): & \mathbf{\dfrac{10a + b}{a+b}} &=& \mathbf{4 + \dfrac{3}{a+b}} \quad | \quad \times (a+b) \\\\ & 10a + b &=& 4(a+b) + 3 \\ & 10a + b &=& 4a+4b + 3 \\ & 10a-4a + b-4b &=& 3 \\ & 6a-3b &=& 3 \quad | \quad : 3 \\ & 2a-b &=& 1 \\ & \mathbf{b} &=& \mathbf{2a-1} \\ \hline \end{array} \)
\(\begin{array}{|lrcll|} \hline & \begin{array}{rcll} \text{We divide a 2-digit positive integer}\\ \text{by the product of its digits} \\ \end{array} \\ \hline (2): & \mathbf{\dfrac{10a + b}{ab}} &=& \mathbf{3 + \dfrac{5}{ab}} \quad | \quad \times (ab) \\\\ & 10a + b &=& 3ab+5 \quad | \quad \mathbf{b=2a-1} \\ & 10a + 2a-1 &=& 3a(2a-1) + 5 \\ & 12a-1 &=& 6a^2-3a + 5 \\ & 6a^2 -15a +6 &=& 0 \quad | \quad : 3 \\ & \mathbf{2a^2 -5a +2} &=&\mathbf{ 0 } \\\\ & a &=& \dfrac{5\pm \sqrt{5^2-4*2*2} }{2*2} \\ & a &=& \dfrac{5\pm \sqrt{9}}{4} \\ & a &=& \dfrac{5\pm 3}{4} \\\\ & a &=& \dfrac{5+ 3}{4} \\\\ & a &=& \dfrac{8}{4} \\\\ & \mathbf{a} &=& \mathbf{2} \\\\ \text{or} & a &=& \dfrac{5- 3}{4} \\\\ & a &=& \dfrac{2}{4} \qquad \text{no integer }! \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline \mathbf{b} &=& \mathbf{2a-1} \quad | \quad \mathbf{a=2} \\\\ b &=& 2*2-1 \\ \mathbf{b} &=& \mathbf{3} \\ \hline \end{array}\)
The 2-digit integer ab is \(\mathbf{23}\)
