# Divisibility by abc.

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#1
Let . Suppose . Does every block of consecutive naturals contain three distinct naturals whose product is divisible by ?

The counter example I have seen to this is with the interval . Is there a smaller/simpler counter example than this, with a lower value of ?
Last edited by I hate maths!; 1 year ago
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1 year ago
#2
(Original post by I hate maths!)
Let . Suppose . Does every block of consecutive naturals contain three distinct naturals whose product is divisible by ?

The counter example I have seen to this is with the interval . Is there a smaller/simpler counter example than this, with a lower value of ?

From a bit of playing:

Just considering the method they used of constructing the numbers based on 3 primes, (7,11,13 in this case), I don't think there is a lower value for c, although a lower inteval should work. With the primes in increasing order, we require , and 7,11,13 is the smallest set that satisfies that.
Last edited by ghostwalker; 1 year ago
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#3
(Original post by ghostwalker)
From a bit of playing:

Just considering the method they used of constructing the numbers based on 3 primes, (7,11,13 in this case), I don't think there is a lower value for c, although a lower inteval should work. With the primes in increasing order, we require , and 7,11,13 is the smallest set that satisfies that.
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1 year ago
#4
(Original post by I hate maths!)
Yes, I neglected to consider the possibliity each prime could go as a square, separately to each number. Which is probably why their solution was based around 6006, rather than 1001.
Last edited by ghostwalker; 1 year ago
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#5
(Original post by ghostwalker)
Yes, I neglected to consider the possibliity each prime could go as a square, separately to each number. Which is probably why their solution was based around 6006, rather than 1001.
I agree with the other bits in your post though, for the reasons you mentioned this probably is the smallest counter example of this form at least (using ). It's much better motivated than I initially thought.
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