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How can we show that between two even natural numbers they're exists a natural number that isn't even?

How can we show that they're exists a natural number that is odd and not divisible by 3, between two multiples of 3 (that are also natural numbers)?

Can we show that they're exists a natural number that is not divisible by any $p_{m}$ less than or equal to $p_{n}$ between two multiples of $p_{n}$ (that are also natural numbers)? I was thinking that maybe we could by applying graph coloring techniques, or perhaps we could rely on modular arithmetic for a proof; or is there another possible method?

What, if any, are the difficulties with this problem?

Brad Graham
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For your first question: if $k$ is the smaller of the two given even numbers, consider $k+1$.

For your second question: if $k$ is the smaller of the two given multiples of $3$, then consider either $k+1$ or $k+2$.

For your last question: the statement is not true. Between $27833$ and $27846$, which are both multiples of $13$, every intermediate number is divisible by one of $2$, $3$, $5$, $7$, or $11$.

Greg Martin
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1.Every even number is of the form $2n$.
Two consecutive even numbers are $2n$ and $2n+2$ so the number you are asking for is $2n+1$ who is clearly odd and between two even numbers.

2.Two consecutive multiples of 3 are $3n$ and $3n+3$ so between the two multiples we have two numbers: $3n+1$ and $3n+2$. One of them is odd and clearly not divisible by 3.
3. If the multiples are $p_n \cdot 1$ and $p_n \cdot 2$ then yes there exists such a number and this is proved by Chebyshev you can find it here. http://en.wikipedia.org/wiki/Proof_of_Bertrand's_postulate
But it has not been proved that what you are asking is true if you consider the general case.
For example we do not know if between ${p_n}^2$ and ${p_n}^2+p_n$
(which are two consecutive multiples of $p_n$ )always exists a prime number.
(a number not divisible by the other $p_m$)
And also it is not true for every multiple of $p_n$ as Greg Martin shows

Konstantinos Gaitanas
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