Can someone please help me with the following problem?
Compute the largest integer power of 6 that divides 73!.
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since there are more multiples of 2 than 3, it should suffice to find the largest power of 3 that divides $73!$. – genisage Oct 28 '14 at 21:49
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@genisage So how many 3 is in $73!$ – CLAUDE Oct 28 '14 at 21:50
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73/3 rounded down + 73/9 rounded down +73/27 rounded down = 24+8+2=34 – Pakquebchsoflwty Oct 28 '14 at 21:51
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Just as many as there are in $72!$ which is two more than there are in $69!$ – genisage Oct 28 '14 at 21:51
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@Pakquebchsoflwty your solution is correct i've checked it with wolframalpha.com. I don't really get why you would take 3, 3^2, and 3^3 though – steffu Oct 28 '14 at 21:58
2 Answers
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You must find the greatest power of $3$ that divides $73!$, since there is much more $2$ in $73!$ rather than $3$. For finding the greatest power of $3$ in $73!$ you can refer to Highest power of a prime $p$ dividing $N!$
So it is equal to $\lfloor\frac{73}{3} \rfloor+\lfloor\frac{73}{9} \rfloor+\lfloor\frac{73}{27} \rfloor=24+8+2=34$.
Hence the greatest power of $6$, that divides $73!$ is $34$
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$$ 73! = 1\cdot2\cdot3\cdot4\cdot5\cdot6\cdots\cdots73. $$ Let's look at how many $3$s from from each of these factors that is a multiple of 3: $$ \begin{array}{ccccc} \text{factor} & \text{3s contributed} \\ 3 & 3 \\ 6 & 3 \\ 9 & 3\cdot3 \\ 12 & 3 \\ 15 & 3 \\ 18 & 3\cdot3 \\ 21 & 3 \\ 24 & 3 \\ 27 & 3\cdot3\cdot3 \\ 30 & 3 \\ 33 & 3 \\ 36 & 3\cdot3 \\ 39 & 3 \\ 42 & 3 \\ \vdots & \vdots \end{array} $$ We get $24$ factors each contibuting at least one $3$.
The ones that are multiples of $9$ contribute at least one more $3$, and there are $8$ of those.
The ones that are multiplies of $27$ contribued at least one more $3$, and there are two of those.
No higher powers of $3$ are involved since we stopped short of $3^4=81$.
So $24+8+2$ is what we're looking for.
(There are more $2$s than $3$s so there are as many $6$s as $3$s.)
