Find the highest power of ## 3 ## dividing ## 823 ##

[Math100]

Homework Statement

Find the highest power of $3$ dividing $823!$.

Relevant Equations

None.

Let $n$ be a positive integer and $p$ be a prime.
Then the exponent of the highest power of $p$ that divides $n!$ is $\sum_{k=1}^{\infty}[\frac{n}{p^{k}}]$.
Observe that $n=823$ and $p=3$.
Thus
\begin{align*}
&[\frac{823}{3}]+[\frac{823}{3^{2}}]+[\frac{823}{3^{3}}]+[\frac{823}{3^{4}}]+[\frac{823}{3^{5}}]+[\frac{823}{3^{6}}]\\
&=274+91+30+10+3+1\\
&=409.\\
\end{align*}
Therefore, the highest power of $3$ dividing $823!$ is $409$.

 

[fresh_42]

Homework Statement:: Find the highest power of $3$ dividing $823!$.
Relevant Equations:: None.

Let $n$ be a positive integer and $p$ be a prime.
Then the exponent of the highest power of $p$ that divides $n!$ is $\sum_{k=1}^{\infty}[\frac{n}{p^{k}}]$.
Observe that $n=823$ and $p=3$.
Thus
\begin{align*}
&[\frac{823}{3}]+[\frac{823}{3^{2}}]+[\frac{823}{3^{3}}]+[\frac{823}{3^{4}}]+[\frac{823}{3^{5}}]+[\frac{823}{3^{6}}]\\
&=274+91+30+10+3+1\\
&=409.\\
\end{align*}
Therefore, the highest power of $3$ dividing $823!$ is $409$.

Looks good.

 

[WWGD]

Sorry, but I suspect you may be doing some double counting here. Multiples of 6,9, etc., are also multiples of 3, etc.
Try the same approach with 10!=3628800, or, easier, with 8!=40320 . By a simple checkup, $2^7$ is the largest power of 2 dividing it.

 

[SammyS]

Sorry, but I suspect you may be doing some double counting here. Multiples of 6,9, etc., are also multiples of 3, etc.
Try the same approach with 10!=3628800, or, easier, with 8!=40320 . By a simple checkup, $2^7$ is the largest power of 2 dividing it.

When @Math100 says that 409 is the highest power of three dividing $823!$, he/she is referring to the exponent used with $3$ as the base, i.e. $\displaystyle 3^{409}$ divides $823!$, but $\displaystyle 3^{410}$ does not.

 

[fresh_42]

When @Math100 says that 409 is the highest power of three dividing $823!$, he/she is referring to the exponent used with $3$ as the base, i.e. $\displaystyle 3^{409}$ divides $823!$.

$3^{409}\,|\,823! \, \wedge \,3^{410}\,\nmid\,823!$

@SammyS : Sorry, it was under way during your correction, and was only meant to correct your typo.

 

[epenguin]

math100 gives a general formula, statement, lemma, theorem, whatever you call it, with no justification or argument. He then says the formula with the two input numbers in question gives another number, which is the answer.

I'd say, whether you think the original statement is obvious or not, he has done nothing really.

 

[SammyS]

math100 gives a general formula, statement, lemma, theorem, whatever you call it, with no justification or argument. He then says the formula with the two input numbers in question gives another number, which is the answer.

I'd say, whether you think the original statement is obvious or not, he has done nothing really.

@Math100 has been going through a course or courses and/or a self study in number theory and related topics. Much of what @Math100 has posted here has been under the excellent guidance of @fresh_42 .

@fresh_42 seemed happy with the the effort in this thread and he's not all that easy to please.

 

[epenguin]

@Math100 has been going through a course or courses and/or a self study in number theory and related topics. Much of what @Math100 has posted here has been under the excellent guidance of @fresh_42 .

@fresh_42 seemed happy with the the effort in this thread and he's not all that easy to please.

I'm aware of Math100´s studies - he has obtained vastly more attention and help here than any other student ever has. I don't think approval by an authority is a criterion that we should be exalting to him (or anyone).

I think his answer is correct but as well as the absence of any argument there seem to be some superficial mistakes:
The sum is not infinite but finite;
The fractions in his first line do not appear to equal the numbers in his second.
If his series is correct he could sum it into more compact expression.

So I don't think this question has been sufficiently answered yet.

 

[SammyS]

Homework Statement:: Find the highest power of $3$ dividing $823!$.
Relevant Equations:: None.

Let $n$ be a positive integer and $p$ be a prime.
Then the exponent of the highest power of $p$ that divides $n!$ is $\sum_{k=1}^{\infty}[\frac{n}{p^{k}}]$.
Observe that $n=823$ and $p=3$.
Thus
\begin{align*}
&[\frac{823}{3}]+[\frac{823}{3^{2}}]+[\frac{823}{3^{3}}]+[\frac{823}{3^{4}}]+[\frac{823}{3^{5}}]+[\frac{823}{3^{6}}]\\
&=274+91+30+10+3+1\\
&=409.\\
\end{align*}
Therefore, the highest power of $3$ dividing $823!$ is $409$.

To clarify:
In the summation you show, I suppose the fraction is meant to mean integer division. It would be good to mention this. Alternatively, use the "floor" function to make this clear. You might not be aware of how to code that in LaTeX: \lfloor \rfloor as in left floor and right floor.

$\displaystyle \quad \sum_{k=1}^{\infty}\left\lfloor\frac{n}{p^{k}}\right\rfloor$

Of course, for any particular values of $n$ and $p$, there are only a finite number of non-zero terms for this infinite sum.

 

[epenguin]

I had been going to say "Some unstated notational meaning for the square brackets?"