Register to reply 
Towers of integers and divisibilty by 11 
Share this thread: 
#1
Feb506, 11:51 PM

P: 5

I'm really stuck on the following problem:
I'm trying to determine whether or not 5^10^5^10^5 is divisible by 11... i have tried a few different methods and can't figure this out. I know the trick must have something to do with modulo 11, but im not sure exactly how to get the result. Any help would be GREATLY appreciated. Thanks in advance. 


#2
Feb506, 11:58 PM

Sci Advisor
HW Helper
P: 944

I can tell you just from looking at it that it's not divisible by 11. Hint: the prime factorization has to include 11 to be divisible by 11, and your number is 5^...



#3
Feb606, 12:13 AM

P: 5

but is there a way to break down the "tower" into a smaller integer using modulo 11?
i know that indeed this integer is not divisible by 11, but many students are confused by the tower. For example, 21 is equivalent to 2 for mod 23. Is there a way to find that 5^10^5^10^5 is equivalent to some integer for mod 11? 


#4
Feb606, 12:23 AM

Sci Advisor
HW Helper
P: 3,147

Towers of integers and divisibilty by 11
5 is raised to a power that is an integer. Therefore, the factorization of your number contains only factors of 5 (a LOT of factors of 5 but only factors of 5). That is equivalent to 0 mod 11.
I don't see the source of confusion. 


#5
Feb606, 12:37 AM

P: 5

Ok, so let's take it one step further:
would 10^5^10^5^10 + 5^10^5^10^5 be divisble by 11? i know that each tower is not divisble by 11, but what about when you add them together? i'm not sure why this is confusing me so much...maybe i'm making it too complicated? 


#6
Feb606, 04:21 AM

Sci Advisor
HW Helper
P: 9,397

What is the order of 5, or ten, mod 11?
Failing that, just work out 5^10 mod 11, then raise that to the power 5, mod it by 11, then rinse and repeat. 10 is easy since it is 1 mod 11, so in your example 10^junk =1 mod 11 of course anything to the power p1 mod p a prime is easy to work out by Fermat't little theorem. 


#7
Feb606, 04:40 PM

P: 5

Here is how i plan to write up the problem:
We notice that powers of 10 (mod 11) form a pattern: 10^0 mod 11 = 1 10^1 mod 11 = 10 10^2 mod 11 = 1 10^3 mod 11 = 10 . . . So all the odd powers of 10 are equal to 10 (mod 11). Clearly, the exponent of the first term is odd: it's some power of 5. Now, we also notice that powers of 5 (mod 11) form a pattern: 5^0 mod 11 = 1 5^1 mod 11 = 5 5^2 mod 11 = 3 5^3 mod 11 = 4 5^4 mod 11 = 9 .............. 5^5 mod 11 = 1, and the above pattern continues. So, if the exponent is a multiple of 5, we'll get a 1. Well, the exponent is a multiple of 10, so we do get a one. Therefore, 10^5^10^5^10 mod 11 = 10 + 5^10^5^10^5 mod 11 = 1 So, 10 + 1 mod 11 = 0. Then we see that this integer is divisible by 11. Ok, that's what I have. It seems right, but some of the wording/explanations seem a little awkward to me. Any comments on this would be greatly appreciated! 


#8
Feb706, 12:24 PM

P: 342

You please see whether you have written what you mean in the post #1.
Do you mean to add the powers or power the powers? 


#9
Feb706, 10:35 PM

P: 5

add the two "towers" of exponents
thats is 5^...... + 10^................ 


#10
Feb806, 01:50 PM

P: 342

addingtwice = multiplying with two.
Your terms is 5^10^5^10, and not something like 5^10 + 10^5 and all such complicated things. The terms can be simply expressed as the product of two primes 5 and 2 with some powers. 


Register to reply 
Related Discussions  
Divisibilty properties  Calculus & Beyond Homework  0  
Number Thry: Divisibilty, perfect square.  Calculus & Beyond Homework  6  
Building tower out of balsa wood that needs to hold a load  Classical Physics  0  
Finding out if 2003^2004  2005 is divisible by 10?  Calculus & Beyond Homework  5  
Telephone Towers  Computing & Technology  2 