On the distribution of multiples of 7 into intervals of length 11

Question

Say we have two primes, say 7 and 11. We are to consider the positions of the multiples of 7 inside the (7 buckets of) multiples of $11$.

So the buckets of 11 are:

$[1,11],[12,22],\ldots ,[67,77]$, 7 in total.

And the multiples of 7 are $7,14,21,28,\ldots,70,77$, 11 in total.

By the PP we can say each bucket contains at least one multiple, but we can't place an upper limit on the contents of each bucket. How can we prove that there are $3$ buckets with 1 multiple and 4 buckets with 2 multiples, and then generalize the result?

Answer 1

Clearly since $7<11$ there is at least one multiple in each bucket and since $2\times 7 = 14 >11$ there are no more than two multiples in each bucket. Since we know the total number of multiples of 7 and the total number of buckets, these limits serve to give the result of $4$ buckets with $2$ multiples in and $3$ buckets with only one multiple in.

When considering $x$ buckets of size $y$, we know that every bucket will contain at least $k=\lfloor y/x \rfloor$ and no more than $\lceil y/x \rceil$ multiples, since $kx\le y$ and $(k+1) x > y$ by definition. The excess after all buckets have $k$ multiples apportioned to them is $(y \bmod x)$. So there will be $(y \bmod x)$ buckets (might be zero) containing $\lceil y/x \rceil$ multiples and the remainder with $\lfloor y/x \rfloor$ multiples in.

Answer 2

Let $p_1$ and $p_2$ be two different numbers (not necessarily prime), and for simplicity's sake assume that they are coprime. Suppose $kp_1 < p_2 < (k+1)p_1$. Each $p_2$-bucket contains at least $k$ multiples of $p_1$, and at most $k+1$ multiples of $p_1$. There are $p_1$ buckets in total, and suppose $n$ of them contain exactly $k$ multiples of $p_1$ ($p_1 - n$ contain exactly $k+1$).

Then there is a unique solution to the equation $nk + (p_1-n)(k+1) = p_2$, so that necessarily we have $n = (k+1)p_1 - p_2$ buckets containing $k$ multiples, and $p_1 - n$ buckets containing $k+1$.

In your case, $k = 1$, $p_1 = 7$ and $p_2 = 11$, so that $n = 2 \times 7 - 11 = 3$ buckets with $k=1$ multiple of $p_1 = 7$, and $p_1 - n = 4$ buckets with $k+1 = 2$ multiples of $7$.

Answer 3

The interval $[11k, 11(k+1)]$ contains $1$ or $2$ multiples of seven depending on $k \space (\mod 7)$ in the following manner:

$$k \equiv 0 : 2$$ $$k\equiv 1 : 2$$ $$k\equiv 2 : 1$$ $$k\equiv 3 : 2$$ $$k\equiv 4 : 1$$ $$k\equiv 5 : 2$$ $$k\equiv 6 : 2$$