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    11 Feb '07 22:41
    True or false:

    If n is an integer, and n squared minus 5 is divisible by an integer m, then the last digit of m cannot be 3 or 7.
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    11 Feb '07 23:47
    Originally posted by David113
    True or false:

    If n is an integer, and n squared minus 5 is divisible by an integer m, then the last digit of m cannot be 3 or 7.
    false, but only in an uncommon circumstance..
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    11 Feb '07 23:55
    Originally posted by David113
    True or false:

    If n is an integer, and n squared minus 5 is divisible by an integer m, then the last digit of m cannot be 3 or 7.
    true...
    for it to work, the last digit of the squared number (after being squared) would have to be a 2 or an 8
    meaning that n must be even (even number times even number equals even number and viceversa for odd numbers)when two numbers are squared, the last digit of the product is determined by the last digit of n, which must be 2 4 6 8 or 0
    these squares are 4, 16, 36, 64, and 0
    since we are concerned about the last digit of these numbers, we can isolate them as:
    4, 6, 6, 4, and 0
    meaning that the square of any number will never have a 2 or 8 as its last digit...
    so the last digit of m will never be 3 or 7 after subtracting 5...
    solved.
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    12 Feb '07 00:202 edits
    Originally posted by rubberjaw30
    true...
    for it to work, the last digit of the squared number (after being squared) would have to be a 2 or an 8
    Why?
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    12 Feb '07 01:21
    Originally posted by David113
    Why?
    because for the last digit to be either a three or a seven, and ur subtracting a 5 to get that number, then by reverse mathematics we can induce that the previous number's last digit must have been a 2 or a 8
  6. Standard memberXanthosNZ
    Cancerous Bus Crash
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    12 Feb '07 01:27
    Originally posted by rubberjaw30
    because for the last digit to be either a three or a seven, and ur subtracting a 5 to get that number, then by reverse mathematics we can induce that the previous number's last digit must have been a 2 or a 8
    Looks like you can't read.
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    12 Feb '07 01:52
    Originally posted by XanthosNZ
    Looks like you can't read.
    i answered his original question and provided an explanation on top...
    i was correct in my explanation...
    so i dont see what u mean i cant read...
    it is a true statement that n squared minus 5 never holds a 3 or a seven as the final digit as long as n meets the definition of an integer...
    puzzle solved, we can go home now
  8. Subscribersonhouse
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    12 Feb '07 01:53
    Originally posted by David113
    True or false:

    If n is an integer, and n squared minus 5 is divisible by an integer m, then the last digit of m cannot be 3 or 7.
    Do you mean "evenly divisible"? With integer result?
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    12 Feb '07 01:561 edit
    Yes, by "divisible" I meant "evenly divisible".
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    12 Feb '07 01:571 edit
    Originally posted by rubberjaw30

    it is a true statement that n squared minus 5 never holds a 3 or a seven as the final digit as long as n meets the definition of an integer...
    That is indeed true, but that was not my question.
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