Integral solutions

Agnishom · 2014-09-02 21:47:47

How many positive integral solutions are there to abcde=1050 and why?

zetafunc · 2014-09-02 22:10:27

What does the unique factorisation theorem tell you?

Agnishom · 2014-09-03 00:07:58

The fundamental theorem of arithmetic? The factorisation of a number is unique except the order in which they occur.

bobbym · 2014-09-03 01:30:52

Hi;

You asked for the code, I assume they can be the same.

d = Divisors[1050];
s = Tuples[d,5];
ans = Select[s,Times@@#==1050&];

Agnishom · 2014-09-03 01:57:56

Never mind, I did this myself:

In[1]:= s = DeleteDuplicates[Map[Sort, Tuples[Divisors[1050], 5]]];

In[2]:= Select[s, Times @@ # == 1050 &]

Out[2]= {{1, 1, 1, 1, 1050}, {1, 1, 1, 2, 525}, {1, 1, 1, 3, 350}, {1,
   1, 1, 5, 210}, {1, 1, 1, 6, 175}, {1, 1, 1, 7, 150}, {1, 1, 1, 10, 
  105}, {1, 1, 1, 14, 75}, {1, 1, 1, 15, 70}, {1, 1, 1, 21, 50}, {1, 
  1, 1, 25, 42}, {1, 1, 1, 30, 35}, {1, 1, 2, 3, 175}, {1, 1, 2, 5, 
  105}, {1, 1, 2, 7, 75}, {1, 1, 2, 15, 35}, {1, 1, 2, 21, 25}, {1, 1,
   3, 5, 70}, {1, 1, 3, 7, 50}, {1, 1, 3, 10, 35}, {1, 1, 3, 14, 
  25}, {1, 1, 5, 5, 42}, {1, 1, 5, 6, 35}, {1, 1, 5, 7, 30}, {1, 1, 5,
   10, 21}, {1, 1, 5, 14, 15}, {1, 1, 6, 7, 25}, {1, 1, 7, 10, 
  15}, {1, 2, 3, 5, 35}, {1, 2, 3, 7, 25}, {1, 2, 5, 5, 21}, {1, 2, 5,
   7, 15}, {1, 3, 5, 5, 14}, {1, 3, 5, 7, 10}, {1, 5, 5, 6, 7}, {2, 3,
   5, 5, 7}}

That is 36 unordered solutions

For ordered solutions:

In[3]:= Length[Permutations[#]] & /@ %

Out[3]= {5, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 60, 60, 60, \
60, 60, 60, 60, 60, 60, 30, 60, 60, 60, 60, 60, 60, 120, 120, 60, \
120, 60, 120, 60, 60}

In[4]:= Total[%]

Out[4]= 1875

Do you have the Kaboobly Doo Analytical Solution, now?

Last edited by Agnishom (2014-09-03 02:00:04)

bobbym · 2014-09-03 02:03:27

How about if a≠b≠c≠d≠e then there are 480.

Agnishom · 2014-09-03 02:06:44

In[9]:= s = Subsets[Divisors[1050], {5}];

In[10]:= Select[s, Times @@ # == 1050 &]

Out[10]= {{1, 2, 3, 5, 35}, {1, 2, 3, 7, 25}, {1, 2, 5, 7, 15}, {1, 3,
   5, 7, 10}}

In[11]:= Length[Permutations[#]] & /@ %

Out[11]= {120, 120, 120, 120}

In[12]:= Total[%]

Out[12]= 480

That is true. How does it help?

bobbym · 2014-09-03 02:07:31

Which do they want?

Agnishom · 2014-09-03 02:10:00

Probably all the ordered pairs of (a,b,c,d,e)

bobbym · 2014-09-03 02:14:32

With them equal or not?

Agnishom · 2014-09-03 02:17:53

Inequality is not necessarily a condition. The answer should be 1875

bobbym · 2014-09-03 02:21:01

The 480 was easier and lends itself to a nice combinatoric answer.

Agnishom · 2014-09-03 02:22:22

Why?

Last edited by Agnishom (2014-09-03 02:23:19)

bobbym · 2014-09-03 02:31:33

Because there are only 4 base solutions and each one can be permuted in 5! ways.

4 x 5! = 480

What about the problem?

Agnishom · 2014-09-03 02:36:09

You can view the edit history?

What about the problem? I have an analytical solution of which I understand nothing.

bobbym · 2014-09-03 02:43:54

I have an analytical solution of which I understand nothing.

That is about right for an analytical solution.

Please post your analytical solution.

Agnishom · 2014-09-03 02:58:42

I do not memorise things I do not understand. I will make another solution.

bobbym · 2014-09-03 03:01:58

I do not memorise things I do not understand. I will make another solution.

I do not agree with that as it violates rule number 1.

I am looking at a solution right now that would involve generating functions.

Agnishom · 2014-09-03 03:09:48

The rule number one of the tautological club is the first rule of it.

I have an excellent solution:

You have a box [2,3,5,5,7]

You have a=b=c=d=e=1

You have to take away a number from the box and multiply it with one of a, b, c or d and repeat until the box is exhausted.

In how many ways can you do this?

The numbers 2,3,7 can be multiplied with any of a,b,c,d,e. So there are 5^3 ways of doing it.

Either you multiply 5^2 to one of the numbers which can be done in 5 ways or you throw the two 5's at two two different variables which can be done in 10 ways.

So you can do it in a total of 5^3(10+5) ways.

That's great!

Now, why does Rakesh think this involves combinations?

Math Is Fun Forum

#1 2014-09-02 21:47:47

Integral solutions

#2 2014-09-02 22:10:27

Re: Integral solutions

#3 2014-09-03 00:07:58

Re: Integral solutions

#4 2014-09-03 01:30:52

Re: Integral solutions

#5 2014-09-03 01:57:56

Re: Integral solutions

#6 2014-09-03 02:03:27

Re: Integral solutions

#7 2014-09-03 02:06:44

Re: Integral solutions

#8 2014-09-03 02:07:31

Re: Integral solutions

#9 2014-09-03 02:10:00

Re: Integral solutions

#10 2014-09-03 02:14:32

Re: Integral solutions

#11 2014-09-03 02:17:53

Re: Integral solutions

#12 2014-09-03 02:21:01

Re: Integral solutions

#13 2014-09-03 02:22:22

Re: Integral solutions

#14 2014-09-03 02:31:33

Re: Integral solutions

#15 2014-09-03 02:36:09

Re: Integral solutions

#16 2014-09-03 02:43:54

Re: Integral solutions

#17 2014-09-03 02:58:42

Re: Integral solutions

#18 2014-09-03 03:01:58

Re: Integral solutions

#19 2014-09-03 03:09:48

Re: Integral solutions

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