The discussion centers on determining which term should be deleted from the product $A=(1!)\times(2!)\times(3!)\times\cdots\times(2016!)$ to make the resulting product $B$ a perfect square. The analysis concludes that removing the term $1008!$ from $A$ results in $B$ being a perfect square. This is derived from the relationship between factorials and even numbers, specifically that $A$ can be expressed as $1008!\times\Box$, where $\Box$ denotes a perfect square.
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[sp]Since $(2n)! = 2n\times (2n-1)!$, it follows that $(2n-1)!\times(2n)! = \bigl((2n-1)!\bigr)^2\times 2n = 2n\times\Box$ (where $\Box$ denotes a perfect square). Then $$\begin{aligned}A &= (1!\times2!)\times(3!\times4!) \times \cdots \times(2015!\times2016!) \\ &= (2\times\Box) \times (4\times \Box) \times \cdots \times(2016\times\Box) \\ &= 2\times 4\times 6\times \cdots \times 2016 \times\Box \\ &= 2^{1008}\times 1008! \times\Box \\ &= (2^{504})^2 \times 1008!\times\Box = 1008!\times\Box.\end{aligned}$$ Thus if $B$ consists of $A$ with $1008!$ removed from the product, then $B$ is a $\Box.$[/sp]Albert said:$A=(1!)\times(2!)\times(3!)\times--------\times(2016!)$
$B$ is obtained by deleting one term from $A$
now decide which term of $A$ should be deleted ,to make $B$ a perfect square
Didn't we just have this posted by a member? Or is that on a different forum? (It's a good question either way.)Albert said:$A=(1!)\times(2!)\times(3!)\times--------\times(2016!)$
$B$ is obtained by deleting one term from $A$
now decide which term of $A$ should be deleted ,to make $B$ a perfect square