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Math Help - 34!

  1. #1
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    34!

    Given that
    34! = 295 232 799 cd9 604 140 847 618 609 643 5ab 000 000,
    determine the digits a, b, c, d.

    so far i know that b = 0 as 34! has a factor of 5^7

    next what i thought of doing was taking the digital root (excluding c d and a) giving me 139 or 4mod9 or 1mod3

    a+c+d must equal 5mod9 and 2mod3 (which sadly are the exact same thing)

    because the number also has a facotor of 11 so taking the alt digital root gives me 21 +d -a -c given d - a -c is 1mod11.

    Okay I'm not totally sure about what to do from here. some help please.

    p.s. if possible could someone reconmend some good reading on the topic of modular arithmetic. thanks
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  2. #2
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    The reason that the alt digital root method works for factors of 11 is that 10\equiv-1\pmod{11}, so that 10x\equiv-x\pmod{11}, 100x\equiv+x\pmod{11} and so on. There is a similar (slightly more elaborate) technique for multiples of 7 or 13, based on the fact that 1001=7×11×13. This means that 1000\equiv-1\pmod7, so that 1000x\equiv-x\pmod7 and 10^6x\equiv+x\pmod7. The same thing is true with 13 in place of 7.

    To apply this technique to 34!, carry out the alt digital root procedure just as in the case of 11, but using groups of three digits rather than just single digits. This tells you that 295–232+799–cd9+604–140+847–618+609–643+5a0 is a multiple of 7 (and also of 13). That gives you two more equations for a, c and d.
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  3. #3
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    sorry i don't totally undersand how you can take the digital root by using groups of three digits for 7 and 13. Could you explain in a bit more detail please.
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  4. #4
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    Quote Originally Posted by bobak View Post
    sorry i don't totally undersand how you can take the digital root by using groups of three digits for 7 and 13. Could you explain in a bit more detail please.
    Basically what it means is that instead of working in units you have to work in hundreds. Split the digits in the number 34! into groups of three and then form an alternating sum (first group minus second group plus third group, and so on). This gives you the expression that I wrote in my previous comment, 295–232+799–cd9+604–140+847–618+609–643+5a0.

    The sum of the positive terms is 295+799+604+847+609+5a0 = 3654 + 10a (because 5a0 means 500+10a). The sum of the negative terms is 232+cd9+140+618+643 = 1642 + 100c + 10d. As I tried to explain in the previous comment, these two quantities should be equal to each other mod 7 (and also mod 13).

    Thus 3654+10a\equiv1642+100c+10d\pmod7,
    3a\equiv 4+2c+3d\pmod7
    . . . . . (because 3654\equiv0\pmod7,\; 10\equiv3\pmod7,\; 1642\equiv4\pmod7,\; 100\equiv2\pmod7,\;),
    \boxed{3a+5c+4d\equiv4\pmod7}.

    Similarly, if we work mod 13,
    3654+10a\equiv1642+100c+10d\pmod{13},
    1+10a\equiv 4+9c+10d\pmod{13}
    . . . . . (because 3654\equiv1\pmod{13},\; 1642\equiv4\pmod{13},\; 100\equiv9\pmod{13},\;),
    \boxed{10a+4c+3d\equiv3\pmod{13}}.

    Together with the previous congruences \boxed{a+c+d\equiv5\pmod9} and \boxed{a+c-d\equiv10\pmod{11}}, you now have four relations for the three unknown digits a, c and d (use three of them to find the solution, and the fourth one to check it!).
    Last edited by Opalg; November 18th 2007 at 11:29 AM.
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  5. #5
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    Alternatively, and perhaps more simply, note that 80,000,000|34!, and hence the last few digits must be 3 520 000 000. Then considering the number modulo 9 and modulo 11 will give the result easily.

    Assuming you're preparing for BMO1, best of luck .
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  6. #6
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    Quote Originally Posted by Simba View Post
    Alternatively, and perhaps more simply, note that 80,000,000|34!, and hence the last few digits must be 3 520 000 000. Then considering the number modulo 9 and modulo 11 will give the result easily.
    Good point, and definitely simpler than my method. What's more, it leads to the same result as mine (after I made a small edit to correct a misprint).

    Yet another method is to use the vast resources of the internet. You can check the answer here!
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