How can I find the multiplicity of a divisor N.

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A123456
A123456 el 18 de Dic. de 2015
Comentada: John D'Errico el 19 de Dic. de 2015
So say N=8, and 8=2^3 how would I get this in Matlab?

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Guillaume
Guillaume el 18 de Dic. de 2015
Editada: Guillaume el 18 de Dic. de 2015
factor(8)
Possibly, your question is not clear.
  2 comentarios
A123456
A123456 el 18 de Dic. de 2015
What I'm trying to do is write a code that will output all the divisors of an integer, say 100. And I also want to find out the multiplicity of each of the divisors of 100. Would you know how to do that?
the cyclist
the cyclist el 18 de Dic. de 2015
factorList = factor(100);
uniqueFactors = unique(factorList);
multiplicity = histcounts(factorList,[uniqueFactors Inf])

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the cyclist
the cyclist el 18 de Dic. de 2015
log2(8)
Logs in base 2 and 10 are available natively. You can get arbitrary bases by using math.
  1 comentario
A123456
A123456 el 18 de Dic. de 2015
What I'm trying to do is write a code that will output all the divisors of an integer, say 100. And I also want to find out the multiplicity of each of the divisors of 100. Would you know how to do that?

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John D'Errico
John D'Errico el 18 de Dic. de 2015
Editada: John D'Errico el 18 de Dic. de 2015
Consider the number N = 8.
Divide N by 2, checking the remainder. The remainder is the non-integer, fractional part in thedivision. If N/2 is an integer, so the remainder is zero, then N is divisible by 2.
Repeat until the remainder is non-zero. So this is just a while loop. Count the number of successful divisions with a zero remainder. (This is a case where we can safely test for an exact zero value of the remainder.)
In the case of N = 8, we have 8/2 = 4. The remainder is zero. So now repeat, 4/2 = 2. Again, the remainder is 0. One more time, and we have 2/2 = 1. The remainder was once more zero.
On the final pass through the loop, we will see that 1/ 2 = 0.5. The remainder was non-zero. We went through the loop 4 times, failing on the 4th time. So the number 8 has exactly 3 factors of 2.
I'll give you a code fragment that embodies the basic idea, although you need to consider how this might need to change if you wish to count the number of factors of 3 a number contains, or some other prime.
CountFacs = 0;
r = 1;
while r ~= 0
N = N/2;
r = rem(N,1);
if r == 0
CountFacs = CountFacs + 1;
end
end
  2 comentarios
A123456
A123456 el 18 de Dic. de 2015
Thanks for that, that's really helpful. Just a quick question, what is CountFacts?
John D'Errico
John D'Errico el 19 de Dic. de 2015
Think about it. Read the code. What did you want to compute? What is the value of CountFacs AFTER the code block terminates?

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