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Can you tell a 7-digit number is divisible by 4 just by its digits? Watch

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    This cropped up in a permutations Q I was doing. For example, if the number ends in an even number, then it must be divisible by 2 and therefore even itself. But what about divisibility by 4? Can you tell by the end/start digits too?
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    yes, because 100 is divisible by four .. therefore everything above the hundreds units is all you have to look at
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    (Original post by a daedalus)
    yes, because 100 is divisible by four .. therefore everything above the hundreds units is all you have to look at
    I'm a bit confused, care to expand? My Q is actually about 7-digit numbers using 1, 1, 2, 3, 3, 4, 6. How do I ensure divisibility by 4 here? With divisibility by 2, I'd just make sure I'd have 2, 4 or 6 as my 7th digit.
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    (Original post by Physics Enemy)
    I'm a bit confused, care to expand? My Q is actually about 7-digit numbers using 1, 1, 2, 3, 3, 4, 6. How do I ensure divisibility by 4 here? With divisibility by 2, I'd just make sure I'd have 2, 4 or 6 as my 7th digit.
    Just make sure that whatever the last two digits are, if treated as a separate number, they are divisible by 4. eg. 12, 32, 64. Say the number was 1123364, the underlined bit must be divisible by 100, and therefore is divisible by 4 as well, so you only need to look at the last two digits.
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    (Original post by Physics Enemy)
    I'm a bit confused, care to expand? My Q is actually about 7-digit numbers using 1, 1, 2, 3, 3, 4, 6. How do I ensure divisibility by 4 here? With divisibility by 2, I'd just make sure I'd have 2, 4 or 6 as my 7th digit.
    He means 100 is divisible by 4 therefore 200 is, 300 is, 10000 is etc. Therefore if you had 423843928492384 for example you would know that it is indeed divisible by 4 because 84 is divisible by 4.
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    You said seven-digit number. Here are three, using the numbers you mentioned (and only ending in an even number as we've established that's necessary):

    1123346
    3146132
    6321134

    As the poster above says, one hundred is divisible by four. So if we take the hundreds units out of the equation:

    1123300
    3146100
    6321100

    These are all divisible by four. So the remainders are:

    46
    32
    34

    And now all you have to do is quickly check if these are divisible by four. Only the second one is, so only the second of the three numbers (3146132) is divisible by four.

    I don't think I've explained it very well, but did the examples help at all?
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    (Original post by Potally_Tissed)
    Just make sure that whatever the last two digits are, if treated as a separate number, they are divisible by 4. eg. 12, 32, 64. Say the number was 1123364, the underlined bit must be divisible by 100, and therefore is divisible by 4 as well, so you only need to look at the last two digits.
    11233 is not divisible by 100 though.

    EDIT: You mean 1123300 ... with the 64 tagged on. I see the point now regarding 100's.
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    (Original post by Physics Enemy)
    11233 is not divisible by 100 though.

    EDIT: You mean 1123300 ... with the 64 tagged on.
    Yeah.
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    (Original post by Physics Enemy)
    11233 is not divisible by 100 though.

    EDIT: You mean 1123300 ... with the 64 tagged on.
    Yeah exactly..
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    (Original post by Jelkin)
    You said seven-digit number. Here are three, using the numbers you mentioned (and only ending in an even number as we've established that's necessary):

    1123346
    3146132
    6321134

    As the poster above says, one hundred is divisible by four. So if we take the hundreds units out of the equation:

    1123300
    3146100
    6321100

    These are all divisible by four. So the remainders are:

    46
    32
    34

    And now all you have to do is quickly check if these are divisible by four. Only the second one is, so only the second of the three numbers (3146132) is divisible by four.

    I don't think I've explained it very well, but did the examples help at all?
    You explained it well.
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    Thanks for all the help guys/girls.
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    (Original post by Physics Enemy)
    I'm a bit confused, care to expand? My Q is actually about 7-digit numbers using 1, 1, 2, 3, 3, 4, 6. How do I ensure divisibility by 4 here? With divisibility by 2, I'd just make sure I'd have 2, 4 or 6 as my 7th digit.
    The four-times table's terminal digit rotates vigesimally rather than decimally, that's the key to understanding the problem.

    4, 8, 12, 16, 20. 24, 28 ...
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    fwliw there's a wiki page of the divisibility rules for 1-20 http://en.wikipedia.org/wiki/Divisib...1.26ndash.3B20

    found it a few weeks ago when I was looking for something else.
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    any number when the last two digits are divisible by 4 is divisible by 4 its how you can tell it will be an olympics year
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    (Original post by mummyperson)
    any number when the last two digits are divisible by 4 is divisible by 4 its how you can tell it will be an olympics year
    The last Olympics was in 2010... 10/4=2.5
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    (Original post by jklmn)
    The last Olympics was in 2010... 10/4=2.5
    Eh? nobody counts the winter olympics.
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    (Original post by jklmn)
    The last Olympics was in 2010... 10/4=2.5
    She said any number (3 digits or longer) with its last 2 digits divisible by 4, will itself be divisible by 4. That doesn't mean a number's divisibility by 4 implies it's final 2 digits are also divisible by 4. We have A => B; not A <=> B.
 
 
 
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