My Math Forum what is the largest power of 2011 that divides a number

 Number Theory Number Theory Math Forum

 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
Global Moderator

Joined: Nov 2006
From: UTC -5

Posts: 16,046
Thanks: 938

Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms
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
Math Team

Joined: Apr 2012

Posts: 1,579
Thanks: 22

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
Global Moderator

Joined: Nov 2006
From: UTC -5

Posts: 16,046
Thanks: 938

Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms
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: 2,778 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
Global Moderator

Joined: Dec 2006

Posts: 19,178
Thanks: 1646

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.

 Tags 2011, divides, largest, number, power

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post usermind Applied Math 11 May 28th, 2012 06:32 AM fortin946 Number Theory 8 April 13th, 2011 07:21 PM tinynerdi Number Theory 1 August 23rd, 2010 01:25 AM Geir Number Theory 1 April 7th, 2009 11:41 PM moemoe11 Algebra 3 September 15th, 2008 12:46 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top