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 May 14th, 2012, 12:29 AM #1 Newbie   Joined: May 2012 Posts: 2 Thanks: 0 what is the largest power of 2011 that divides a number what is the largest power of 2011 that divides $2010^{2011^{2012}}+2012^{2011^{2010}}$
 May 14th, 2012, 05:09 PM #2 Math Team   Joined: Apr 2012 Posts: 1,579 Thanks: 22 Re: what is the largest power of 2011 that divides a number Ok, I'm voting for the 0th power. 2010 is -1 mod 2011 2012 is +1 mod 2011 If n = -1 mod x, n to any even power = +1 mod x If n = +1 mod x, n to any higher power whatsoever will be +1 mod x Both 2010 and 2011 are raised to higher and specifically even powers. So the sum of the terms will be +2 mod 2011, hence not divisible by 2011. If you are not divisible by x, you can't possibly be divisible by a higher power of x. That's how I see it? Am I missing something? Last edited by skipjack; March 22nd, 2015 at 10:43 AM.
May 14th, 2012, 06:05 PM   #3
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Re: what is the largest power of 2011 that divides a number

Quote:
 Originally Posted by johnr Am I missing something?
The exponents are odd, not even. So the answer is larger than 0 -- actually quite a bit larger.

May 15th, 2012, 04:28 AM   #4
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Re: what is the largest power of 2011 that divides a number

Quote:
Originally Posted by CRGreathouse
Quote:
 Originally Posted by johnr Am I missing something?
The exponents are odd, not even. So the answer is larger than 0 -- actually quite a bit larger.
They are even numbers raised to an odd exponent raised to an even exponent. n^x^y = n^(x*y), no?

Last edited by skipjack; March 22nd, 2015 at 10:47 AM.

May 15th, 2012, 04:52 AM   #5
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Re: what is the largest power of 2011 that divides a number

Quote:
Originally Posted by johnr
Quote:
Originally Posted by CRGreathouse
Quote:
 Originally Posted by johnr Am I missing something?
The exponents are odd, not even. So the answer is larger than 0 -- actually quite a bit larger.
They are even numbers raised to an odd exponent raised to an even exponent. n^x^y = n^(x*y), no?
By convention, exponentiation is right-associative, so n^x^y means n^(x^y) rather than (n^x)^y = n^(x*y).

Last edited by skipjack; March 22nd, 2015 at 10:47 AM.

 May 15th, 2012, 08:38 AM #6 Math Team   Joined: Apr 2012 Posts: 1,579 Thanks: 22 Re: what is the largest power of 2011 that divides a number I see! That does change things, no? But while goofing around with some much smaller examples, I noted that, eg, 2^5 + 4^5 is of course divisible by 3, but is not divisible by any higher power of 3. So I'm curious to see how one would go about trying to solve this question. Brute computation is out of the question for the example given!
 May 16th, 2012, 04:23 AM #7 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: what is the largest power of 2011 that divides a number Perhaps binomial expansion could get us somewhere. Let 2011^n be the requested power. $2010^{2011^2012} + 2012^{2011^2010}= (2011 - 1)^{2011^2012} + (2011 + 1)^{2011^{2010}} \\ \\ = \sum_{k = 0}^{2011^{2012}} {2011^{2012} \choose k} \cdot 2011^k \cdot (-1)^{2011^{2012} - k} + \sum_{k = 0}^{2011^{2010}} {2011^{2010} \choose k} \cdot 2011^k \cdot 1^{2011^{2010}-k} \\ \\ = \sum_{k = 0}^{2011^{2010}} \left( {2011^{2010} \choose k} \cdot 2011^k + {2011^{2012} \choose k} \cdot 2011^k \cdot (-1)^{1- k} \right) + \sum_{k = 2011^{2010}+1}^{2011^{2012}} {2011^{2012} \choose k} \cdot 2011^k \cdot (-1)^{2011^{2012} - k}$ Now I dont know if $\sum_{k= 0}^{2011^{2010}} \left( {2011^{2010} \choose k} \cdot 2011^k + {2011^{2012} \choose k} \cdot 2011^k \cdot (-1)^{1- k} \right) + \sum_{k = 2011^{2010}+1}^{2011^{2012}} {2011^{2012} \choose k} \cdot 2011^k \cdot (-1)^{2011^{2012} - k} \equiv \sum_{k = 0}^{2011^{2010}} \left( {2011^{2010} \choose k} \cdot 2011^k + {2011^{2012} \choose k} \cdot 2011^k \cdot (-1)^{1- k} \right) (\bmod 2011^n)$ i.e. $\sum_{k= 2011^{2010}+1}^{2011^{2012}} {2011^{2012} \choose k} \cdot 2011^k \cdot (-1)^{2011^{2012} - k} \equiv 0 (\bmod 2011^n)$ And if yes, how to proceed. Any ideas?
March 22nd, 2015, 10:46 AM   #8
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Quote:
 Originally Posted by johnr They are even numbers raised to an odd exponent raised to an even exponent. n^x^y = n^(x*y)
Yes, and an odd value raised to any integer power is not even.

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