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My motivation for this question comes from the proof here: https://math.stackexchange.com/a/1734600

The proof states:

For a positive integer n, whenever you divide n by one of its prime factors p, you obtain then number of positive integers â‰Īn which are a multiple of p...

I've worked through a few examples to convince myself it holds, but I'm wondering why it's true in general. Is there a proof for this fact?

When I tried to prove it, I got to:

$\frac n {p_1} = p_1^{a_1 - 1} \cdot (p_2^{a_2} \cdot \ldots \cdot p_k^{a_k})$, assuming that the prime factorization of $n$ is $p_1^{a_1} \cdot \ldots \cdot p_k^{a_k}$. But I'm not sure how I can proceed from here.

Bill Dubuque
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    Let $n = pk$, then the multiples of $p$ that are $\le n$ are $p, 2p, \dots, kp = n$ and there are clearly $k = n/p$ of these. – Hans Engler Apr 27 '21 at 18:14
  • That's basically the definition of division. It's true for all numbers. Not just primes and the reason it is true is ... well, If $n\div q = k$ then you count to $q$ $k$ times before you get to $n$ and each of those $k$ times you count is one multiple of $q$. – fleablood Apr 27 '21 at 22:36
  • Let $\frac np = m$ so $mp = n$. The multiples of $p$ that are less than or equal to $n$ are $p, 2p, 3p, 4p,......,(m-1)p$ and $mp = n$. – fleablood Apr 27 '21 at 22:41

3 Answers3

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You divide $n$ by $p$ to get $k$. So $n = pk$.

Now look at $0\cdot p, 1\cdot p, \ldots, (k-1)\cdot p$.

All those are less than $n$ (why?) and there are $k$ of them. So there are at least $k$ numbers less than $n$ divisible by $p$.

But any natural number divisible by $p$ can be written as $p \cdot b$ for some natural number $b$, and in order for $pb$ to be less than $n$, we must have $b < k$, hence it must be one of the ones listed above.

In short: those $k$ numbers are exactly the natural numbers less than $n$ that are divisible by $p$.

John Hughes
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The numbers $\le n $ which are multiples of $p $ are precisely the numbers $p,2p,3p,\dots,(n/p)p $.

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This is true of all numbers. Not just primes, and it is basically the definition of division.

For any positive integer $n$ and an positive divisor integer $q$ there is a unique integer $k$ so that $kq \le n < (k+1) q$ and $n = kq + r$ were $0 \le r < q$ and $r$ is the "remainder" and $k$ is the quotient.

And $q, 2,q, 3q,......, kq$ are all the multiples of $q$ that are less or equal to $n$. There are $k$ of them.

That's all there is to it.

fleablood
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