Problem 29 · 2014 Math Kangaroo
Stretch
Algebra & Patterns
custom-operationsubstitution
The mapping \(f:\mathbb{Z} o\mathbb{Z}\) fulfils the conditions \(f(4)=6\) and \(xf(x)=(x-3)f(x+1)\). What is the value of the expression \(f(4) imes f(7) imes f(10) imes\ldots imes f(2011) imes f(2014)\)?
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Answer: D — \(2013!\)
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Hint 1 of 2
The relation x·f(x) = (x−3)·f(x+1) lets you step f from one integer to the next.
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Hint 2 of 2
When you multiply the wanted terms, look for a massive telescoping cancellation.
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Approach: use the recurrence and telescope the product
- The condition gives f(x+1) = x·f(x)/(x−3), so with f(4)=6 every value of f is determined.
- Forming f(4)·f(7)·f(10)·…·f(2014) and simplifying, the fractions telescope.
- The product collapses to 2013!.
Mark:
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