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Math Help - Finding prime numbers - problem w/method

  1. #1
    Junior Member Euclid Alexandria's Avatar
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    Question Finding prime numbers - problem w/method

    I suspect I'm just drawing a blank here, but I thought any number is divisible by 3 if the sum of its digits is divisible by 3.

    Am I wrong about this? I came across the number 6111. The sum of its digits is 9. But when I divide 6111 by 3, I get 2033\frac{1}{3}.
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    Quote Originally Posted by Euclid Alexandria
    I suspect I'm just drawing a blank here, but I thought any number is divisible by 3 if the sum of its digits is divisible by 3.

    Am I wrong about this? I came across the number 6111. The sum of its digits is 9. But when I divide 6111 by 3, I get 2033\frac{1}{3}.
    6111 divided by 3 is 2037
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    Member Glaysher's Avatar
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    Quote Originally Posted by Euclid Alexandria
    I suspect I'm just drawing a blank here, but I thought any number is divisible by 3 if the sum of its digits is divisible by 3.

    Am I wrong about this? I came across the number 6111. The sum of its digits is 9. But when I divide 6111 by 3, I get 2033\frac{1}{3}.
    This is true and as PerfectHacker said you didn't divide 6111 correctly.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Glaysher
    This is true and as PerfectHacker said you didn't divide 6111 correctly.
    Note as well, if the sum of the digits is divisible by 9 it is also divisible by 9.

    -Dan
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    Quote Originally Posted by topsquark
    Note as well, if the sum of the digits is divisible by 9 it is also divisible by 9.

    -Dan
    And if the last two digits are divisible by 4 than the whole number is divisible by 4
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    Forum Admin topsquark's Avatar
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    Euclid Alexandria

    I have an idea of what you are up to, but I would like to ask...I am curious as to what your post has to do with finding prime numbers?

    -Dan
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    Quote Originally Posted by topsquark
    Euclid Alexandria

    I have an idea of what you are up to, but I would like to ask...I am curious as to what your post has to do with finding prime numbers?

    -Dan
    There is a absolute method to determine if a number is a prime or not. It is just not very useful even for super-computers.
    If and only if,
    \left[ \frac{(n-1)!+1}{n} \right] = \frac{(n-1)!+1}{n}
    Of course that is the greatest integer function.
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    Junior Member Euclid Alexandria's Avatar
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    Quote Originally Posted by topsquark
    Euclid Alexandria

    I have an idea of what you are up to, but I would like to ask...I am curious as to what your post has to do with finding prime numbers?

    -Dan
    Good question. I was actually part-way through simplifying a fraction when I came across 6111. While subtracting/carrying in the division problem, I suavely made the 1 into a 10 instead of 11, hence 2033\frac{1}{3}

    So yeah, it was really a problem w/division
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    Quote Originally Posted by Euclid Alexandria
    Good question. I was actually part-way through simplifying a fraction when I came across 6111. While subtracting/carrying in the division problem, I suavely made the 1 into a 10 instead of 11, hence 2033\frac{1}{3}

    So yeah, it was really a problem w/division
    It's always nice to see someone do work out by hand (even if they mess up)
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    Junior Member Euclid Alexandria's Avatar
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    I'm avoiding my calculator at all costs right now. Of course, it seems I'm regressing lately. Moving on to 2037, the sum of its digits is 12. I just tried dividing that by 2, and then tried again, and both times I got 1018\frac{1}{2}

    Shouldn't 2 be going evenly into 2037?
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    Quote Originally Posted by Euclid Alexandria
    I'm avoiding my calculator at all costs right now. Of course, it seems I'm regressing lately. Moving on to 2037, the sum of its digits is 12. I just tried dividing that by 2, and then tried again, and both times I got 1018\frac{1}{2}

    Shouldn't 2 be going evenly into 2037?
    No, 2 only goes into even numbers...

    Finding the sum of a numbers digits only works for 3 and 9
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    Junior Member Euclid Alexandria's Avatar
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    Smile

    Yeah, it hit me last night that 2 only applies to numbers that end in 0, 2, 4, 6 and 8. Like I said, regressing!
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    Junior Member Euclid Alexandria's Avatar
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    Question

    Ok, is this correct? Something doesn't look right with this.

    \frac{9,506}{12,222}
    = \frac{4,753 \cdot \not2}{679 \cdot \not2 \cdot 3 \cdot 3}
    = \frac{4,753}{6,111}
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    Quote Originally Posted by Euclid Alexandria
    Ok, is this correct? Something doesn't look right with this.

    \frac{9,506}{12,222}
    = \frac{4,753 \cdot \not2}{679 \cdot \not2 \cdot 3 \cdot 3}
    = \frac{4,753}{6,111}
    It can be a lot simpler than that:

    It simplifies down to =\frac{7}{9}

    Highest Common Factor of the pair is 1358, calculators are very useful for this sort of simplifying.
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    Junior Member Euclid Alexandria's Avatar
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    Question How to determine the Highest Common Factor?

    Quote Originally Posted by Random333 View Post
    It can be a lot simpler than that:

    It simplifies down to =\frac{7}{9}

    Highest Common Factor of the pair is 1358, calculators are very useful for this sort of simplifying.
    I was browsing through my subscribed threads for problems that were never completely dealt with. This is one such thread that I'd like to understand. In response to Random's answer (or anyone else reading this), both my book and my previous instructor encouraged the use of calculators for checking answers, but discouraged their use for solving problems. I agree with the idea of not allowing the calculator to do any of my work for me (at this point in time).

    By "the pair," I suppose you're referring to \frac{9,506}{12,222}

    And not \frac{4,753}{6,111}

    Since 4,753 is a prime number.

    Given the restraint of solving by hand, I'm having difficulty determining how I might ascertain the pair's Highest Common Factor... especially if this is the only method available for simplifying this fraction in particular.
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