Divisibilty


by Virtate
Tags: divisibilty
Virtate
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#1
Oct7-05, 09:40 AM
P: 6
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.
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Tom Mattson
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#2
Oct7-05, 09:55 AM
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We do help people with homework here, but you have to show how you started and where you got stuck.
Virtate
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#3
Oct7-05, 10:00 AM
P: 6
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.

Tom Mattson
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#4
Oct7-05, 10:07 AM
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Divisibilty


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.
Tom Mattson
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#5
Oct7-05, 10:11 AM
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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.
Virtate
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#6
Oct7-05, 10:50 AM
P: 6
Thanks Tom!


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