\(\large \begin{cases}{ \color{blue}a \times \color{brown}b \times \color{green}c = \color{violet}{15}} \\ {\color{brown}b \times \color{green}c\times \color{red}d = \color{violet}{30}} \\ {\color{green}c\times \color{red}d \times \color{blue}a = \color{violet}{10}} \\ {\color{red}d \times \color{blue}a \times \color{brown}b = \color{violet}{6}}\end{cases} \)

Given a, b, c, d are four distinct natural numbers that satisfy the system of equations above.

Determine the value of a+b+c+d.

Guest Jul 5, 2020

#2**0 **

Here's an alternative method.

\(\displaystyle abc = 15, \text{ so, }abcd=15d,\text{ so, }\frac{abcd}{15}=d \quad \dots\quad(1).\)

Similarly,

\(\displaystyle \frac{abcd}{30}=a,\quad \dots\quad (2),\\ \displaystyle \frac{abcd}{10}=b,\quad \dots \quad(3),\\ \displaystyle \frac{abcd}{6}=c,\quad \dots \quad (4).\)

Adding (1), (2), (3) and (4),

\(\displaystyle a+b+c+d=abcd\left(\frac{1}{15}+\frac{1}{30}+\frac{1}{10}+\frac{1}{6}\right)=abcd\left(\frac{11}{30}\right)\quad \dots \quad (5). \)

Multiplying the original four equations,

\(\displaystyle a^{3}b^{3}c^{3}d^{3}=15\times 30\times10\times6=2^{3}\times3^{3}\times5^{3},\\ \text{ so }abcd=2\times 3 \times 5 = 30.\)

Then, from (5),

\(\displaystyle a+b+c+d=30.\left(\frac{11}{30}\right)=11.\)

Guest Jul 5, 2020