I hate maths!
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Let a,b,c \in \mathbb{N}. Suppose a<b<c. Does every block of c consecutive naturals contain three distinct naturals whose product is divisible by abc?

The counter example I have seen to this is a=77, b=91, c=143 with the interval 5930,5931,...,6072. Is there a smaller/simpler counter example than this, with a lower value of c?
Last edited by I hate maths!; 1 year ago
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ghostwalker
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(Original post by I hate maths!)
Let a,b,c \in \mathbb{N}. Suppose a<b<c. Does every block of c consecutive naturals contain three distinct naturals whose product is divisible by abc?

The counter example I have seen to this is a=77, b=91, c=143 with the interval 5930,5931,...,6072. Is there a smaller/simpler counter example than this, with a lower value of c?

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 925, 926,...,1067 should work. With the primes in increasing order, we require 2p_1>p_3, and 7,11,13 is the smallest set that satisfies that.
Last edited by ghostwalker; 1 year ago
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I hate maths!
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(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 925, 926,...,1067 should work. With the primes in increasing order, we require 2p_1>p_3, and 7,11,13 is the smallest set that satisfies that.
I am unsure about that interval. What about 968 \times 980 \times 1014=(7^2 \times 11^2 \times 13^2) \times 960?
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ghostwalker
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(Original post by I hate maths!)
I am unsure about that interval. What about 968 \times 980 \times 1014=(7^2 \times 11^2 \times 13^2) \times 960?
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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I hate maths!
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(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 p_1, p_2, p_3). It's much better motivated than I initially thought.
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