Student Support Forum: 'programming help, finding factors of LCM' topicStudent Support Forum > General > "programming help, finding factors of LCM"
Author 
Comment/Response 
JD

07/08/13 5:07pm
Okay, so once again, I have read the documentation and the manual,
I do exactly what it says to do, and mathematica either will not work or repeats what I just said.
It's supposed to be world changing software, but I can't get it to calculate anything.
Does it need gas?
This is not homework, but it is critical to my research. And my future, so Sum [ " Thank You ", { i, 1, Infinity } ] for your help.
So here's what I am doing.
I'm working on finding the number of factors that a number will have.
This number is an LCM of consecutive integers that are greater than 0.
So the LCM of 1 through 6 is ( 1 x 2 x 3 x 2 x 5 x 1 ), which equals 60.
The prime factors of 60 are 2,3,2,and 5. This is ( 2^2 x 3^1 x 5^1 ).
If we examine the exponents or powers of the prime factors, there is a 2, a 1, and a 1.
If I add 1 to each exponent, giving a 3, a 2, and a 2, and multiply those numbers together,
it will yield the number of factors. There are ( 3 x 2 x 2 ) = 12 factors of 60.
This count includes 1 and the number itself as factors :
( 1 , 60 ), ( 2 , 30 ), ( 3 , 20 ), ( 4 , 15 ), ( 5 , 12 ), ( 6 , 10 ) .
1 2 3 4 5 6 7 8 9 10 11 12 .
It's an interesting trick to know and it works on any number once the prime factors have been determined.
I have tried using :

In[26]:= Product[Floor[1 + log_n [6], {n \[Element] Primes && n \[Element] Integers && n >= 2 && n <= 6}]]
During evaluation of In[26]:=
Product::argmu : "\[NoBreak]Product\[NoBreak] called with 1 \
argument; \[NoBreak]2\[NoBreak] or more arguments are expected. \
\!\(\*ButtonBox["\[RightSkeleton]",
Appearance>{Automatic, None},
BaseStyle>"Link",
ButtonData:>"paclet:ref/message/General/argmu",
ButtonNote>"Product::argmu"]\)"
Out[26]= Product[Floor[1 + log_n[6], {n \[Element] Primes && n \[Element] Integers && n >= 2 && n <= 6}]]
In[28]:= Sum[log[Floor[1 + log_n [6], {n \[Element] Primes && n \[Element] Integers && n >= 2 && n <= 6}]]]
During evaluation of In[28]:=
Sum::argmu : "\[NoBreak]Sum\[NoBreak] called with 1 argument; \
\[NoBreak]2\[NoBreak] or more arguments are expected. \
\!\(\*ButtonBox["\[RightSkeleton]",
Appearance>{Automatic, None},
BaseStyle>"Link",
ButtonData:>"paclet:ref/message/General/argmu",
ButtonNote>"Sum::argmu"]\)"
Out[28]= Sum[log[Floor[1 + log_n[6], {n \[Element] Primes && n \[Element] Integers && n >= 2 && n <= 6}]]]
In[21]:= FactorInteger [60]
Out[21]= {{2, 2}, {3, 1}, {5, 1}}

A couple of things are obviously not working out.
I want the log in the base of the primes up to 6, of the number 6.
Not 6 multiplied by the log_p of nothing to the right.
I'm new to using logs, so if this is correct then that's amazing.
So I want to see a result more like : Floor[1+log_2(6)] x Floor[1+log_3(6) x Floor[1+log_5(6) = 12
And if I were to simply use FactorInteger, it gives the prime factors and their exponents in {}
but doesn't really do the math for me.
I'm actually going to be working with much larger LCM's once i have the basics down.
I will actually only be concerned with the number of digits of the number of factors.
So, later, instead of finding the Product, I will want to use " Sum[log_10[Floor[1+log_p(n)]] ,
where n is the last consecutive interger of the LCM (1,2,3,...,n),
and p is a prime greater or equal to 1 and p is less than or equal to n .
I figured that I could just skip 1 because Log_10(1) = 0
since this will not change the sum
and I didn't want to confuse the computer with trying to find the exponent of 1
and also because 1 is technically not a prime,
I could just as well exclude it.
But it is counted as a factor, along with the number itself,
using the method with the exponents even though an exponent for 1 is never given.
It works so I guess I shouldn't dwell on it.
If the elders will not grant me their great wisdom,
I am determined to pay to get this done.
URL: , 

   
 
