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Subject: "modulus arithmetic"     Previous Topic | Next Topic
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Conferences The CTK Exchange College math Topic #469
Reading Topic #469
4878bayley
Member since Aug-19-04
Aug-19-04, 08:06 PM (EST)
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"modulus arithmetic"
 
   hi and good morning

i am a new member. i have read a few of the recent messages and they all seem to contain information at a more advanced level than mine.

i have returned to study after quite a while. although i am ok with most calculations i do have problems with questions such as:-

eg find 5^1000 X 3^1000 in arithmetic mod 8 (ie find five to the power of one-thousand multiplied by three to the power of one-thousand)

i have worked out the modulus 8 table but nothing springs to mind in terms of an answer to the above.

i also am having problems with :-

eg what will the last digit of 2^2001 be? explain why this digit is the same as 2^2001 in arithmetic mod 10.

so far i have spent around three weeks trying to find out the answers, reading books and searching on the internet and also asking friends. please could anyone help. i need to understand how to do these and how to think about the problem.

thank you
kind regards
mel


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Cino Hilliard
guest
Aug-25-04, 06:23 AM (EST)
 
1. "RE: modulus arithmetic"
In response to message #0
 
   >hi and good morning
>
>eg find 5^1000 X 3^1000 in arithmetic mod 8 (ie find five
>to the power of one-thousand multiplied by three to the
>power of one-thousand)
You can reduce some multiplications by re-writing this as
(15^40)^25
base 8 this is the
166515747...543666601 1303 digit number.
base 10 it is the
12338405...962890625 1177 digit number.

>i have worked out the modulus 8 table but nothing springs to
>mind in terms of an answer to the above.
>
>i also am having problems with :-
>
>eg what will the last digit of 2^2001 be? explain why this
>digit is the same as 2^2001 in arithmetic mod 10.

The last digit of 2^2001 base 8 is 0. The actual nunmber base 8 is
10000...00000 This has 2001*log(2)/log(8) + 1 = 668 digits.

Base 10 the number is
22962...058752 This has 2001*log(2)/log(10) + 1 = 603 digits.
Base 2 of course has 2002 digits.

You can get ideas on how to solve this by building a table of powers
into the teens and note that even powers end in 4 or 6 and odd powers
end in 2 or 8. Then you will note 6 occurs when the power is
divisible by 4, 2^4 = 16, 2^8 = 256, 2^12 = 4096 etc. if 4 does not
divide the power then 4 is the last digit. For odd powers, notice 2
occurs when power is 1,5,9,13,17. See the pattern? These numbers are
of the form 4n+1. so 2001 = 500*4 + 1 and so 2^2001 ends in 2 as is
shown above.

>
>so far i have spent around three weeks trying to find out
>the answers, reading books and searching on the internet and
A good book is Elementary number Theory by Jones and Jones. Amazon.com
has it. Another great help will be downloading the Pari Calculator
to compute these numbers and test your theories.

https://pari.math.u-bordeaux.fr/

Download this as suggested by the installer. TThis will give you
some of the fancy options.

Check the documentation files to learn Pari.
https://pari.math.u-bordeaux.fr/doc.html

If you have windows xp, download the Pari.exe executable and install
it. You will not be able to do the base conversions as is because
that is a special script I wrote. you can get this script at

https://groups.yahoo.com/group/B2LCC/files/Pari/

Download the util.gp and .gprc and place in your working directory.
I sugggest c:\pari\gpfiles

>thank you
>kind regards
>mel

Have fun
CLH


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sfwc
Member since Jun-19-03
Oct-01-04, 12:57 PM (EST)
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2. "RE: modulus arithmetic"
In response to message #1
 
   >You can reduce some multiplications by re-writing this as
>(15^40)^25
>base 8 this is the
>166515747...543666601 1303 digit number.
>base 10 it is the
>12338405...962890625 1177 digit number.
That doesn't really help with the calculation unless use of a computer is allowed.

Try 3^1000 * 5 ^1000 =(mod 8) (3*5)^1000
=(mod 8) 15^1000
=(mod 8) (-1)^1000
=(mod 8) 1^500
=(mod 8) 1

Alternatively you could note that 3^1000 * 5^1000 is both odd and square and then check in your multiplication table that any odd square must be congruent to 1 mod 8.

>The last digit of 2^2001 base 8 is 0. The actual nunmber
>base 8 is
>10000...00000 This has 2001*log(2)/log(8) + 1 = 668 digits.
>
>Base 10 the number is
>22962...058752 This has 2001*log(2)/log(10) + 1 = 603
>digits.
>Base 2 of course has 2002 digits.
>
>You can get ideas on how to solve this by building a table
>of powers
>into the teens and note that even powers end in 4 or 6 and
>odd powers
>end in 2 or 8. Then you will note 6 occurs when the power is
>divisible by 4, 2^4 = 16, 2^8 = 256, 2^12 = 4096 etc. if 4
>does not
>divide the power then 4 is the last digit. For odd powers,
>notice 2
>occurs when power is 1,5,9,13,17. See the pattern? These
>numbers are
>of the form 4n+1. so 2001 = 500*4 + 1 and so 2^2001 ends in
>2 as is
>shown above.

Alas, you have not actually provided a proof. Such 'pattern spotting' attacks often work but they do not guarantee a correct answer. It's a better idea to use something along these lines:

To find 2^2001 mod 10, We first note that as 2^2001 is even we need only find 2^2001 mod 5 (why?)

Now we wish to find a small n with 2^n =(mod 5) 1 (You'll see why in a minute). A brief calculation shows that n = 4 will do. Noting that 2001 = 4*500 + 1 we can finish off:

2^(2001) =(mod 5) 2^(4*500) * 2
=(mod 5) (2^4)^(500) * 2
=(mod 5) 1^500 * 2
=(mod 5) 1 * 2
=(mod 5) 2

so it is either 2 or 7 mod 10 and so (since it is even) must be 2 mod 10.

I also note that you assume, rather than prove, that this will be the same as the last digit in base 10. The proof is fiddly but not too hard so I will leave it for you to have fun working it out.

Thankyou

sfwc
<><


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