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  1. #1
    Uranium Lounger
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    I probabaly have something better to do (2003 115612.pii)

    I just ran a Find and Replace to get the column S references reset to column U (I didn't want to carry over formatting) and created the following formula:

    =IF(U60,1-uUM(U31:U45,U52)/U60,)

    It doesn't error out even though =uUM() isn't valid function. It should return '#NAME?' (as it does when I remove the IF part of the formula. Can anyone reproduce this? Have I missed a flaw in Excel?
    -John ... I float in liquid gardens
    UTC -7ąDS

  2. #2
    Platinum Lounger
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    Re: I probabaly have something better to do (2003

    Just tried it and the formula only produces a #NAME error if cell U60 is true

  3. #3
    New Lounger
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    Re: I probabaly have something better to do (2003 115612.pii)

    John,

    have you tried it with U60 = True and U60 = false ?

    Andrew

  4. #4
    Uranium Lounger
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    Doh

    You nailed it. I never reailzed that the error doesn't appear until the IF returns TRUE; I expected the syntax error to be highlighted regardless.
    -John ... I float in liquid gardens
    UTC -7ąDS

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