Can the Product of n Consecutive Positive Integers Be Divisible by n!?

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SUMMARY

The discussion centers on proving that the product of n consecutive positive integers is always divisible by n! using the minimal counterexample technique. The approach involves assuming the statement is false and identifying the least counterexample k, which leads to the conclusion that the proposition holds for k-1. Algebraic manipulation is required to demonstrate that if the proposition is true for k-1, it must also be true for k. The definition of divisibility is clarified, emphasizing the role of the integer c in the context of the proof.

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Homework Statement


Using the minimal counterexample technique prove that the product of n consecutive positive integers is always divisible by n!

The Attempt at a Solution



Suppose that the statement is not true and the product of n consecutive positive integers is not always divisible by n!...

I'm stuck here. How can I continue this to get a contradiction and thus show that the product of is always divisible by n! ? Any guidance is greatly appreciated.
 
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Then there is a least counterexample k for which the proposition does not hold. Since this is the least counterexample, the proposition holds for k-1. Now you need to do some algebraic tricks to show that the proposition holding for k-1 implies that it holds for k. Your set up should look something like
c*(k-1)!= a * (a+1) *...*(n+a-1), for some a,c>0
 
JThompson said:
Then there is a least counterexample k for which the proposition does not hold. Since this is the least counterexample, the proposition holds for k-1. Now you need to do some algebraic tricks to show that the proposition holding for k-1 implies that it holds for k. Your set up should look something like
c*(k-1)!= a * (a+1) *...*(n+a-1), for some a,c>0

What does that "c" mean? Could you please explain a bit more how you got this expression?
 
Note: Sorry- the last term in the equation I posted had an error- the last term should have been a+k-2.

Definition of "divides": For integers x and y, where x does not equal 0, we say x divides y if there exists an integer c so that y=xc.

So in the equation c*(k-1)!= a * (a+1) *...*(a+k-2), the terms a, a+1, a+2 ... a+k-2 are the consecutive integers (note that there are a+k-2 - a +1 = k-1 of them), and c is the integer from the definition of integer divisibility above. This is the inductive assumption.

Edit: Forget the part about the inductive assumption- different type of proof. Forgot this one used least counterexample.
 

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