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    Amyunimus's Avatar
    Amyunimus Posts: 11, Reputation: 1
    New Member
     
    #1

    Jul 5, 2006, 11:39 PM
    Finding and Listing Value in Cells
    Question:

    In 10 cells in a row, two unspecified values will appear. In all other cells, there is a zero. For example:

    A B C D E F G H I J K
    0 61 0 0 28 0 0 0 0 0 0

    I have many rows, and the column in which they appear is randomized. In each row, one of the two values falls in the "correct" location. Let's say here, I know the number in column "B" is correct.

    Using this information, I can list "61" in a separate column. But is there any way to find the second number, "28" without having the additional information of where it is located?

    Thanks.
    ScottGem's Avatar
    ScottGem Posts: 64,966, Reputation: 6056
    Computer Expert and Renaissance Man
     
    #2

    Jul 6, 2006, 07:23 AM
    I'm not sure I follow what you want. But there are some things that you might be able to do.

    Lets say you want to just show the second value in a separate column. The formula:

    =Sum(A1:K1) - B1 Will return 28

    You could also create a custom function that loops through each column in the row and determines which has a non zero value.
    colbtech's Avatar
    colbtech Posts: 748, Reputation: 66
    Senior Member
     
    #3

    Jul 6, 2006, 09:02 AM
    Can you give us more info on what you are trying to achieve.

    It's easy enough to display the values if they are non-zero, but??
    kp2171's Avatar
    kp2171 Posts: 5,318, Reputation: 1612
    Uber Member
     
    #4

    Jul 6, 2006, 09:12 AM
    if the values are truly randomized and you have many rows then you'd see the pattern for both the "proper" position for both values quite quickly (with just a few rows), as you state that each row must have one of the two numbers in the right place.
    Amyunimus's Avatar
    Amyunimus Posts: 11, Reputation: 1
    New Member
     
    #5

    Jul 6, 2006, 08:17 PM
    Thanks-- summing the rows and then subtracting the correct number works in this case (ScottGem's suggestion). Thank you!
    ScottGem's Avatar
    ScottGem Posts: 64,966, Reputation: 6056
    Computer Expert and Renaissance Man
     
    #6

    Jul 7, 2006, 06:45 AM
    Glad to assist

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