Three friends

Three friends had balls in ratio 2: 7: 4 at the start of the game. Could they have the same number of balls at the end of the game? Write 0, if not, or write the minimum number of balls they had together.

Correct result:

n =  39

Solution:

$$\begin{gathered} n=LCM(3,2+7+4)=39 \\ a={\displaystyle \frac{2}{2+7+4}}\cdot n={\displaystyle \frac{2}{2+7+4}}\cdot 39=6 \\ b={\displaystyle \frac{7}{2+7+4}}\cdot n={\displaystyle \frac{7}{2+7+4}}\cdot 39=21 \\ c={\displaystyle \frac{4}{2+7+4}}\cdot n={\displaystyle \frac{4}{2+7+4}}\cdot 39=12 \\ d=n\mathrm{/}3=39\mathrm{/}3=13 \\ n=39 \end{gathered}$$

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