Items tagged with number_theory

Hi, given a, b and n such that a^x is congurent to b (mod n) how could one write a procedure to calculate x using baby step giant step method to calculate x for example i would like to be able to show that x=60 when a=3 b=64 and n=137

The 196 algorithm goes like this.  Start with an integer.  Reverse the digits.  Add the reversed number to the integer.  For most numbers, this eventually leads to a palendrome.  That is to say the number is equal to the reversed number.  I wrote a little Maple procedure to explore 196, the smallest number that will probrably never become a palendrome when put into the algorithm.

 

Let me know if you like my code.

Regards,
Matt

proc4.pdf

proc4.mw

http://mathworld.wolfram.com/196-Algorithm.html

 

Ex: Give: Sum = 16;

Result: [1,3,5,7]

we have positive number from 1 to 1000. how many time we write number 3?

A square has 36 sub-squares in it. How to Number each sub squares from 1 to 36, to make the sum of vertical, the sum of horizontal, and the sum of cross line are the same .Describe in general if possible.

  Can I dowload the Iterator package and use it with Maple 12? Thanks.
 I obtained a count of 7 with the command: stirling2(4,2) . How do I get a list of the 7 partitions? Thanks.

Recently I saw this question (http://www.mapleprimes.com/posts/141668-Partitions-Of-A-Natural-Number-Into-Factors). I want to find the divisors of a number, say 120, :

2,4,6,8,10,12,15,20,24,30,40,60,120

for the students but couldn't find a simple way for that expect the solutions in above link. Isn't an easier way to do this job in Maple?

Thanks so for the time

Recently, I encountered the problem in which one's asked to investigate if the following fraction is an integer or not. Of course the problem has an theoritical approach, but I wanted to make the problem dynamic. The fraction is p=(k^2-87)/(3k+117) and we want to find k in which p is an integer. What I did is to use the function msolve and a loop for but it gives my just the empty set which is incorrect. Any hint? Thanks

Find all the positive integers n ,such that

Prove that (k^{{2}^{n}}-1)\equiv 0 (mod2^{n+2}).If

If (abc‾) and (abcabcabc...abc‾)Ξ0(mod91).Find all non-negative integers (a,b,c)?

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