The number of positive integral solutions is to be found

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Discussion Overview

The discussion revolves around finding the number of positive integral solutions to the equation xyz=3000. Participants explore various mathematical approaches, including combinatorial methods and inequalities, to solve the problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents the equation xyz=3000 and requests assistance in finding positive integral solutions.
  • Another participant suggests factoring 3000 into primes and framing the problem as a combinatorics question.
  • A different participant shares their attempts at finding solutions using specific values for x and proposes using the AM-GM inequality and the multinomial theorem.
  • One participant expresses concern about overthinking the problem and introduces a simpler analogy involving allocating colored balls into buckets.
  • Another participant provides a formula for the answer, indicating a potential solution to the problem.
  • A later reply raises the question of finding all integral solutions, suggesting a shift in the scope of the discussion.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the methods for solving the problem, and multiple approaches and interpretations are presented without resolution.

Contextual Notes

Some participants express uncertainty about the application of specific mathematical techniques, and there is a lack of clarity regarding the assumptions behind the proposed methods.

Who May Find This Useful

This discussion may be useful for students preparing for entrance exams, particularly those interested in combinatorial mathematics and problem-solving strategies.

ack
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I am preparing for an entrance and I came across this sum.The equation given is xyz=3000. we need to find how many positive integral solutions are there for x,y and z.
Please help.
 
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Let's factor 3000 into primes.
We are left with 23*3*53.

Now we are left with a very simple combinatorics question. Can you figure it out? Think about what items you are selecting and what you are placing them into.
 
I already tried that.x can be 2^0,2^1,2^2or2^3 that is 4 ways ,2ways for 3and 4ways for 5.But what about y?
I thought of another method ,
using AM>GM,

x+y+z>=43 and max can be 3002 (when one of them is 3000 nd the other two 1 each.)
Then use multinomial theorem.Can this be done?
 
ack said:
I already tried that.x can be 2^0,2^1,2^2or2^3 that is 4 ways ,2ways for 3and 4ways for 5.But what about y?
I thought of another method ,
using AM>GM,

x+y+z>=43 and max can be 3002 (when one of them is 3000 nd the other two 1 each.)
Then use multinomial theorem.Can this be done?
You are vastly overthinking this.
Consider the following problem: we have 3 red balls, 3 blue balls, and a green ball. How many different ways can we allocate them between 3 buckets?
 
Oh!...then the answer would be
9P7/(3!)^2.
thanks a lot!
 
Oh!...then the answer would be
9P7/(3!)^2.
thanks a lot!
 
Supposing we were to find all integral solutions?
 

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