towers of integers and divisibilty by 11

by scottyk
Tags: divisibilty, integers, towers
scottyk is offline
Feb5-06, 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.
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Alkatran is offline
Feb5-06, 11:58 PM
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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^...
scottyk is offline
Feb6-06, 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?

Tide is offline
Feb6-06, 12:23 AM
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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.
scottyk is offline
Feb6-06, 12:37 AM
P: 5
Ok, so let's take it one step further:

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?
matt grime
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Feb6-06, 04:21 AM
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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 p-1 mod p a prime is easy to work out by Fermat't little theorem.
scottyk is offline
Feb6-06, 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!
vaishakh is offline
Feb7-06, 12:24 PM
P: 343
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?
scottyk is offline
Feb7-06, 10:35 PM
P: 5
add the two "towers" of exponents

thats is 5^...... + 10^................
vaishakh is offline
Feb8-06, 01:50 PM
P: 343
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.

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