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Let $S(a)$ be a set of the multiples of an integer $a$.

Then, here is my question.

Question : Is the following true for any $n\ge 2\in\mathbb N$ ?

Supposing that any two of $n$ integers $a_1, a_2,\cdots,a_n$ are coprime, then there exists a set of $n$ consective integers $k,k+1,\cdots,k+n-1$ such that $$k+i\in S(a_j)\ \ \text{($i=0,1,\cdots,n-1$)}.$$ Suppose that if $k+p\in S(a_r), k+q\in S(a_s)$ and $p\not=q$ , then $r\not=s$.

Example : Suppose that $a_b$ represents that $a$ is a multiple of $b$.

The $(n,a_1,a_2,a_3)=(3,3,5,7)$ case has an example $[48_3, 49_7, 50_5]$.

The $(n,a_1,a_2,a_3,a_4)=(4,2,3,5,11)$ case has an example $[8_2,9_3,10_5,11_{11}]$.

The $(n,a_1,a_2,a_3,a_4,a_5)=(5,2,3,5,7,11)$ case has an example $[119_7,120_5,121_{11},122_2,123_3]$.

Motivation : We know the $n=2$ case is true. I've been able to prove that the $n=3,4$ cases are true. The above proposition seems true for $n$ in general, but I'm facing difficulty for proving that. Can anyone help?

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up vote 1 down vote accepted

We will prove a slightly stronger statement.

Let $n \geq 2$ be an integer, and $a_1,\ldots,a_n$ pairwise coprime integers. Then there exists an integer $k$ such that $k + i \in S(a_{i+1})$ for all $0 \leq i \leq n-1$.

The $n=2$ case is obvious. We will use induction on $n$. Let $n > 2$, and let $a_1,\ldots,a_n$ be pairwise coprime integers. By induction hypothesis, there exist integers $k,l$ such that $k + i \in S(a_{i+1})$ for all $0 \leq i \leq n-2$ and $l + i \in S(a_{i+2})$ for all $0 \leq i \leq n-2$. Write $A = a_1 a_2 \cdots a_{n-1}$, $B = a_2 a_3 \cdots a_n$ and $C = a_2 a_3 \cdots a_{n-1}$. For any $0 \leq i \leq n-3$ we have $l + i \in S(a_{i+2})$ and $k + 1 + i \in S(a_{i+2})$, and thus $l - (k+1)$ is divisible by every $a_j$ for $2 \leq j \leq n-1$. So $l - (k-1)$ is also divisible by $C$. So we can write $l - (k-1) = xC$ for some integer $x$.

Now note that gcd$(A,B) = C$, and thus there exist integers $s,t$ such that $sA + tB = C$. We see that $l-(k+1) = xC = xsA + xtB$, which we can write as $l - xtB = (k+1) + xsA$. Now consider the consecutive integers $k + xsA, k + xsA + 1, \ldots, k + xsA + (n-1)$. For $0 \leq i \leq n-2$, $A \in S(a_{i+1})$, and thus $k + xsA + i \in S(a_{i+1})$. Finally, we note that $k + xsA + (n-1) = l - xtB + (n-2)$. As $B \in S(a_n)$, we get $k + xsA + (n-1) \in S(a_n)$. So the integers $k + xsA, k + xsA + 1, \ldots, k + xsA + (n-1)$ satisfy the conditions, which completes the proof.

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A nice and simple answer! Thank you very much. – mathlove Nov 20 '13 at 16:03

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