The greatest common divisor of two integers is (x+2) and their least common multiple is x(x+2), where x is a positive integer. If one of the integers is 24, what is the smallest possible value of the other one?
+26367 The greatest common divisor of two integers is \((x+2)\)
and their least common multiple is \(x(x+2)\),
where \(x\) is a positive integer.
If one of the integers is \(24\),
what is the smallest possible value of the other one ?
Let \(b\) is the value of the other one
greatest common divisor\((24,b) = x+2 \text{ and } x \in N\)
least common multiple\((24,b) = x(x+2) \text{ and } x \in N\)
Formula:
\(\begin{array}{|rcll|} \hline \text{greatest common divisor}(a,b) \times \text{least common multiple}(a,b) = a\times b \\ \hline \end{array}\)
\(\mathbf{b =\ ?}\)
\(\begin{array}{|rcll|} \hline 24\times b &=& (x+2)\times x(x+2) \\\\ \mathbf{b} & \mathbf{=} & \mathbf{\dfrac{x(x+2)^2}{24}} \qquad x \in N, ~ x \gt 0 \\ \hline \end{array} \)
\(\begin{array}{|r|c|c|} \hline x \in N & \mathbf{b=\dfrac{x(x+2)^2}{24}} & \text{integer}\ ? \\ \hline 1 & \dfrac{3}{8} \\\\ 2 & \dfrac{4}{3} \\\\ 3 & \dfrac{25}{8} \\\\ 4 & \color{red}6 & \checkmark \\\\ \cdots & \cdots \\\\ \hline \end{array}\)
The smallest possible value of the other one is \(\mathbf{6}\) and \(x = 4\)
Check:
\(\begin{array}{lcl} \text{greatest common divisor}(24,6) &=& 4+2 = 6 \\ \text{least common multiple}(24,6) &=& 4(4+2)= 4\cdot 6 = 24 \\ \end{array}\)
The smallest value of the other integer appears to be 60.
GCD[24, 60] =12, so (x + 2) =12 and x=10
LCM[24, 60] =120, so that x(x + 2) =10 x 12 = 120
+26367 The greatest common divisor of two integers is \((x+2)\)
and their least common multiple is \(x(x+2)\),
where \(x\) is a positive integer.
If one of the integers is \(24\),
what is the smallest possible value of the other one ?
Let \(b\) is the value of the other one
greatest common divisor\((24,b) = x+2 \text{ and } x \in N\)
least common multiple\((24,b) = x(x+2) \text{ and } x \in N\)
Formula:
\(\begin{array}{|rcll|} \hline \text{greatest common divisor}(a,b) \times \text{least common multiple}(a,b) = a\times b \\ \hline \end{array}\)
\(\mathbf{b =\ ?}\)
\(\begin{array}{|rcll|} \hline 24\times b &=& (x+2)\times x(x+2) \\\\ \mathbf{b} & \mathbf{=} & \mathbf{\dfrac{x(x+2)^2}{24}} \qquad x \in N, ~ x \gt 0 \\ \hline \end{array} \)
\(\begin{array}{|r|c|c|} \hline x \in N & \mathbf{b=\dfrac{x(x+2)^2}{24}} & \text{integer}\ ? \\ \hline 1 & \dfrac{3}{8} \\\\ 2 & \dfrac{4}{3} \\\\ 3 & \dfrac{25}{8} \\\\ 4 & \color{red}6 & \checkmark \\\\ \cdots & \cdots \\\\ \hline \end{array}\)
The smallest possible value of the other one is \(\mathbf{6}\) and \(x = 4\)
Check:
\(\begin{array}{lcl} \text{greatest common divisor}(24,6) &=& 4+2 = 6 \\ \text{least common multiple}(24,6) &=& 4(4+2)= 4\cdot 6 = 24 \\ \end{array}\)
