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Finding out if 2003^2004 - 2005 is divisible by 10?

  1. Oct 7, 2005 #1
    Hello,

    I was wondering, how does one go finding out if 2003^2004 - 2005 is divisible by 10??? Or that 3^102 * 7^29 is divisible by 33???

    If someone could help me, I would really appreciate it.

    Thank You.
     
  2. jcsd
  3. Oct 7, 2005 #2

    Tom Mattson

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    We do help people with homework here, but you have to show how you started and where you got stuck.
     
  4. Oct 7, 2005 #3
    I only got as far as 2003 = 3 (mod 10), and I have no idea where to go from there. I think I'm headed in the wrong direction... If someone could give me a few pointers, that would be great.
     
  5. Oct 7, 2005 #4

    Tom Mattson

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    OK, in order for 2003^2004-2005 to be divisible by 10, it has to end in zero. That means that 2003^2004 has to end with 5 (because when you subtract 2005, you'll get a zero in the ones place). Now you should be able to tell pretty straighforwardly if 2003^2004 ends with 5.

    As for the other one, any number that is divisible by 33 must be divisible by both 3 and 11. Since you were given the prime factorization of the other number, you should be able to tell just by looking whether it is divisible by 33.
     
  6. Oct 7, 2005 #5

    Tom Mattson

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    Incidentally, it was me who moved your thread to the K-12 HW section. I can see from your second thread that this is probably a College course, so I've moved both this thread and your other one to College HW.

    Any and all homework questions go to this area, not in the Math section.
     
  7. Oct 7, 2005 #6
    Thanks Tom!
     
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