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# NumTheSL3#39

0
261
1 No calc please

Jun 8, 2019

#1
+3

Let's say it starts wih  x  fish.

 number of fish left after John leaves _ $$=\ \frac23({\color{Aquamarine}x}-1)\ =\ \frac23x-\frac23$$ number of fish left after Joe leaves $$=\ \frac23(\ {\color{Aquamarine}\frac23x-\frac23}\ -1)\ =\ \frac49x-\frac{10}{9}$$ number of fish left after _James_ leaves $$=\ \frac23(\ {\color{Aquamarine}\frac49x-\frac{10}{9}}\ -1)\ =\ \frac{8}{27}x-\frac{38}{27}$$

The smallest positive integer value of  x  that makes  $$\frac{8}{27}x-\frac{38}{27}$$  a positive integer is  25 .

By looking at a graph we can check this:  https://www.desmos.com/calculator/mivvh3t6in

So the minimum possible number of fish before John threw out the first fish is  25 .

Jun 8, 2019

#1
+3

Let's say it starts wih  x  fish.

 number of fish left after John leaves _ $$=\ \frac23({\color{Aquamarine}x}-1)\ =\ \frac23x-\frac23$$ number of fish left after Joe leaves $$=\ \frac23(\ {\color{Aquamarine}\frac23x-\frac23}\ -1)\ =\ \frac49x-\frac{10}{9}$$ number of fish left after _James_ leaves $$=\ \frac23(\ {\color{Aquamarine}\frac49x-\frac{10}{9}}\ -1)\ =\ \frac{8}{27}x-\frac{38}{27}$$

The smallest positive integer value of  x  that makes  $$\frac{8}{27}x-\frac{38}{27}$$  a positive integer is  25 .

By looking at a graph we can check this:  https://www.desmos.com/calculator/mivvh3t6in

So the minimum possible number of fish before John threw out the first fish is  25 .

hectictar Jun 8, 2019