Thread: Find the first lowest value in a column that exceeds $foo
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Ron Rosenfeld Ron Rosenfeld is offline
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Default Find the first lowest value in a column that exceeds $foo
On Sat, 21 Mar 2009 21:44:28 -0400, Ron Rosenfeld
wrote:

=1/FLOOR(1/D1*100000000,1)*100000000

Simplified:

=100000000/INT(100000000/D1)


--ron
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