+0

# GCD/LCM stuffness

0
105
3

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?

Guest Aug 29, 2018

#3
+20105
+3

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:

$$\begin{array}{lcl} \text{greatest common divisor}(24,6) &=& 4+2 = 6 \\ \text{least common multiple}(24,6) &=& 4(4+2)= 4\cdot 6 = 24 \\ \end{array}$$

heureka  Aug 30, 2018
#1
0

Guest Aug 29, 2018
#2
+1

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

Guest Aug 30, 2018
edited by Guest  Aug 30, 2018
#3
+20105
+3

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:

$$\begin{array}{lcl} \text{greatest common divisor}(24,6) &=& 4+2 = 6 \\ \text{least common multiple}(24,6) &=& 4(4+2)= 4\cdot 6 = 24 \\ \end{array}$$

heureka  Aug 30, 2018