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I came across this problem while studying for a midterm and I'm having a tough time proving the first implication:

Let $n\in\Bbb Z$. Prove that $n$ is a multiple of both $5$ and $9$ if and only if it is a multiple of $45.$

I know that:

If n is a multiple of $45, n = 45a,$ where $a \in\Bbb Z.$ Then, $n = 5(9a) = 5b$, where $b = 9a$. Thus, $n$ is a multiple of $5$.

Likewise, $n = 45a = 9(5a) = 9c$, where $c = 5a$. Thus, n is also a multiple of $9$.

Where I have trouble is proving that if $5$ and $9$ divide $n$, $45$ also divides $n.$

$n = 5a$ and $n = 9b$, where $a,b \in\Bbb Z.$

I'm pretty lost as to what my next step here should be. Anyone have a hint to point me in the right direction?

Shaun
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nicons
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    Hint: $5$ and $9$ are coprime. – Shaun Feb 20 '20 at 23:02
  • $n=5a$ and $n=9b$ so your next step should be noting ..... $5a = 9b$.... What does that say? Well, for one thing $a =\frac {9b}5$ is an integer. (And $b = \frac {5a}9$ is an integer). So .....? – fleablood Feb 21 '20 at 00:23
  • Alternatively: To brainstorm: You know that $n=5a$and $n=9b$ and you know that you are going to have $n = 45k$. For the sake of bouncing ideas off a wall what do you think $k$ is going to be in terms of $a$ and $b$. ($n = 9a = 5b = 45k$ so ... $a =5k$ and $b = 9b$....) Now how can we prove this must happen? – fleablood Feb 21 '20 at 00:27

3 Answers3

1

HINT

You have $5a = n = 9b$ so $5a = 9b$. Since $5$ and $9$ are relatively prime,

  • $9|(5a)$ so either $9|5$ or $9|a$
  • $5|(9b)$ so either $5|9$ or $5|b$

Can you complete this?

gt6989b
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Note $1=2\cdot 5-9$. Imagine $5\mid n$ and $9\mid n$. Then: $n=5u=9v$. Now we have:

$$n=5u=5(2\cdot 5u-9u)=5(2\cdot 9v-9u)=45(2v-u)$$

and so $45\mid n$.

0

Let $n=\varepsilon p_1^{\alpha_1}p_2^{\alpha_2}\dots p_m^{\alpha_m}$ be the prime decomposition of $n$, where $p_i$ are ordered with $p_1$ the smallest prime and $p_m$, the largest, and $\alpha_i\in\Bbb N$ and $\varepsilon=\pm 1$.

Since $5\mid n$, we have $n=5a$ for some $a\in \Bbb Z$. Similarly, $n=9b$, some $b\in\Bbb Z$. But $3^2=9$.

Therefore, by uniqueness in the Fundamental Theorem of Arithmetic, we have $5a=n=3^2b=\varepsilon p_1^{\alpha_1}p_2^{\alpha_2}\dots p_m^{\alpha_m}$ implies that both $5$ and $3^2$ divide the decomposition, so, if $n$ is even, $p_1=2$ and $\alpha_1\ge 1$ and we have $p_2=3$ with $\alpha_2\ge 2$ and $p_3=5$ with $\alpha_3\ge 1$. Hence $3^2\times 5=45\mid n$.

Can you do the case when $n$ is odd?

Shaun
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