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Largest Integer Divisible by All Integers


Date: 01/01/2002 at 00:43:23
From: Salman
Subject: Number theory

Hello, Dr. Maths,

Please help me in solving the following problem:

Show that 24 is the largest integer divisible by all integers less 
than its square root.

Thanks.


Date: 01/02/2002 at 15:28:04
From: Doctor Rob
Subject: Re: Number theory

Thanks for writing to Ask Dr. Math, Salman.

Any such number n must satisfy the following:  Let m be the greatest 
integer such that m^2 <= n.  Then

  LCM(1,2,3,...,m) | n.

But when n >= 25, LCM(1,2,3,...,m) > n, so this division is 
impossible. This is the crux of the matter, and the part that requires 
proof.

When n >= 25, we get that m >= 5.

Now Bertrand's Postulate says that for every real number x > 1 there 
is a prime number p with x < p < 2*x. Thus there is a prime number p 
with m < p < 2*m,

   LCM(1,2,...,2*m) >= LCM(1,2,...,p),  because p < 2*m,
                    >= LCM(1,2,...,m)*p,  because p > m and
                                                    p is prime,
                    > n*p,  by hypothesis on n,
                    > n*m,  because p > m,
                    >= 5*n,  because m >= 5,
                    >= 5*m^2,  because n >= m^2,
                    > 4*m^2,  because m > 0,
                    = (2*m)^2.

This eliminates all numbers from n to n*p >= 7*n.  Use this fact 
inductively to eliminate all n >= 25.

Feel free to write again if I can help further.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Number Theory

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