Dear MathCentral,

A man has three goldfish. When the youngest goldfish was born, the oldest fish was three times the middle fish's age. Nine years ago the oldest fish's age was the sum of the two other fish's ages. How old are the three goldfish?

Sincerely,
Nathan

Hi Nathan,

Suppose the youngest fish is X years old, the middle fish is Y years old and the oldest fish is Z years old. I drew up a table of the ages of the fish.

DateYoungest FishMiddle FishOldest Fish
NowXYZ
Nine years agoX - 9Y - 9Z - 9 = (X - 9)+(Y - 9)
X years agoX - XY - XZ - X = 3(Y - X)

The first row is now when the ages are X, Y and Z. The second row is 9 years ago when the ages were X - 9, Y - 9 and Z - 9. At this time the age of the oldest is the sum of the ages of the other two, that is Z - 9 = (X - 9)+(Y - 9). The last row is when the youngest fish was born, that is X years ago.

From the last two rows of the table

Z = X + Y - 9 and
Z = 3Y - 2X
Thus 2Y = 3(X - 3)

X and Y are integers and 2 divides the left side of the equation above, hence 2 must divide the right side. This forces X to be odd. We also know that X is larger than 9 and hence X can be 11, 13, 15...

Try X = 11, solve 2Y = 3(X - 3) for Y and then substitute into Z = 3Y - 2X to find Z. Do the three values satisfy the required conditions? Now try X = 13 and X = 15...

I hope this helps,
Penny
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