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

[Virtate]

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.

 

[quantumdude]

We do help people with homework here, but you have to show how you started and where you got stuck.

 

[Virtate]

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.

 

[quantumdude]

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.

 

[quantumdude]

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]

Thanks Tom!