MHB What term should be deleted to make a perfect square?

[Albert1]

$A=(1!)\times(2!)\times(3!)\times--------\times(2016!)$
$B$ is obtained by deleting one term from $A$
now decide which term of $A$ should be deleted ,to make $B$ a perfect square

 

[Opalg]

$A=(1!)\times(2!)\times(3!)\times--------\times(2016!)$
$B$ is obtained by deleting one term from $A$
now decide which term of $A$ should be deleted ,to make $B$ a perfect square

[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]

 

[topsquark]

$A=(1!)\times(2!)\times(3!)\times--------\times(2016!)$
$B$ 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.)

-Dan