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  1. #1
    Bronze Lounger
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    IF begins with (2002)

    =IF(K3="","",IF(LEFT(K3="446280"),"XYZ"),"")

    I'm getting #VALUE

  2. #2
    Plutonium Lounger
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    Mar 2002
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    Re: IF begins with (2002)

    LEFT(K3="446280") makes no sense. You must specify how many characters you want within the LEFT function: LEFT(K3,6)

    The formula becomes

    =IF(LEFT(K3,6)="446280","XYZ","")

    Note that it isn't necessary to test for K3="" separately.

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