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    Joined: Jan 2009
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    I am a very big fan of symmetry and found this kind of neat. I know folks here won't think I'm a complete freak for finding this interesting. Hopefully, it's not so obvious that the rest of you say, DUH!!! And maybe one of you mathy folk can tell me whether there's an explanation for these patterns.

    So, here's my random question....

    A while back when DS was working on his multiplication math facts, I noticed some patterns, and wonder why they exist. Please forgive me if I get the math terms wrong. I think the examples will explain.

    Example One: As far as I can tell, the sum of the digits of any product with a factor of 9 will always be 9.
    ~ 9 x 6 = 54, and 5 + 4 = 9
    ~ 9 x 481 = 4329, and 4 + 3 + 2 + 9 = 18, and 1 + 8 = 9

    Example Two: Similiar effect for products with a factor of 6, except that the sum of the digits is always either 3, 6 or 9.
    ~ 6 x 782 = 4692, and 4 + 6 + 9 + 2 = 21, and 2 + 1 = 3

    Other numbers have different patterns,
    Products of factors of 8: the sum of the digits go down by one with the next factor. Of course, they have to skip zero....
    ~ 8 x 1 = 8
    ~ 8 x 2 = 16, and 1 + 6 = 7
    ~ 8 x 3 = 24, and 2 + 4 = 6
    ~ 8 x 4 = 32, and 3 + 2 = 5
    ~ 8 x 5 = 40, and 4 + 0 = 4

    Factors of 7: the sum of the digits go down by two for each subsequent factor.
    ~ 7 x 1 = 7
    ~ 7 x 2 = 14, and 1 + 4 = 5
    ~ 7 x 3 = 21, and 2 + 1 = 3
    ~ 7 x 4 = 28, and 2 + 8 = 10, and 1 + 0 = 1
    ~ 7 x 5 = 35, and 3 + 5 = 8
    ~ 7 x 6 = 42, and 4 + 2 = 6

    Anyway, thanks for reading the long post. I was just wondering if there is an explanation for it. If not, writing this should help me stop pondering it. smile


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    You'll probably like this site:
    http://www.dr-mikes-math-games-for-kids.com/6-times-table-tips.html
    It doesn't really explain why but it does have more of the fun patterns.

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    Thanks for sharing, Inky! Looks like a great site!

    (Hmmm, I wonder how many resources you're allowed to suggest to a given teacher before you are considered a pain in the neck. Maybe I'll get DS interested in what's on the site and HE can suggest it to his teacher.)

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    Originally Posted by Mama22Gs
    Example One: As far as I can tell, the sum of the digits of any product with a factor of 9 will always be 9.

    Will always be divisible by 9 itself, or will sum to 9 if you keep summing the digits. (So 999 sums to 27, which sums to 9, so the original number is divisible by 9, as are the intermediate steps.) Very useful for accountants, because a transposition error is always divisible by 9, so if I'm off by $27, it means I probably transposed digits.

    Related: A number divisible by 3 has digits that sum to a number divisible by 3.

    Originally Posted by Mama22Gs
    Example Two: Similiar effect for products with a factor of 6, except that the sum of the digits is always either 3, 6 or 9.

    Side-effect of the general "divisible by 3" rule. Anything divisible by 6 is by definition also divisible by 3.

    Originally Posted by Mama22Gs
    I was just wondering if there is an explanation for it. If not, writing this should help me stop pondering it. smile

    Yes, and my recollection is that the explanations are pretty simple, particularly for the "change in the digits" ones.

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    Yes, number theory and modular mathematics directly address this, and there are some great proofs on-line and in "popular-math" books (and in some pretty math-intensive books on the subject). Most of these disciplines deal with more advanced theory (prime numbers, divisibility, Chinese Remainder Theorem...), but they are all routed in patterns like this.

    Last edited by LilMick; 11/10/10 11:49 AM.
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    I can prove the 9 rule, although it's hard to type out.

    Let n be a positive integer such that the sum of the digits is divisible by 9. We can write the digits of n as dk, dk-1, dk-2, ..., d1, d0. (Supposed to be subscripts) So d0+d1+d2+...+dk=9m for some integer m. We can write out the decimal expansion of n as 10^k*dk+10^(k-1)*dk-1+...+10d1+d0. Separate this sum into [(999...9)dk+(999...9)dk-1+...99d2+9d1]+[d0+d1+d2+...+dk]. The first sum is divisible by 9, so we can equate it to 9s. The second sum is 9m (from above). So we have n=9s+9m=9(s+m). Hence, 9 divides n.

    I hope this helps. You can prove a lot of the divisibility rules this way. I teach a discrete math course for gifted hs students. A great resource for number theory is the Epp discrete math book.

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    Thanks, PinkPanther.

    It may take me a while to digest what you've written, but I will give it a go. smile


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