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Find the maximum power of $24$ in $(48!)^2$ ?

How to approach for such questions ?

Jon Garrick
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3 Answers3

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$24=\color\red{2^3}\cdot\color\green{3^1}$

The multiplicity of $\color\red{2}$ in $48!$ is $\sum\limits_{n=1}^{\log_{\color\red{2}}48}\Big\lfloor\frac{48}{\color\red{2}^n}\Big\rfloor=24+12+6+3+1=\color\red{46}$

The multiplicity of $\color\green{3}$ in $48!$ is $\sum\limits_{n=1}^{\log_{\color\green{3}}48}\Big\lfloor\frac{48}{\color\green{3}^n}\Big\rfloor=16+5+1=\color\green{22}$

Therefore:

  • The maximum value of $n$ such that $(\color\red{2^3})^n$ divides $48!$ is $\Big\lfloor\frac{\color\red{46}}{\color\red{3}}\Big\rfloor=\color\red{15}$
  • The maximum value of $n$ such that $(\color\green{3^1})^n$ divides $48!$ is $\Big\lfloor\frac{\color\green{22}}{\color\green{1}}\Big\rfloor=\color\green{22}$

Therefore:

  • The maximum value of $n$ such that $(\color\red{2^3})^n$ divides $(48!)^2$ is $\Big\lfloor\frac{\color\red{46}\cdot2}{\color\red{3}}\Big\rfloor=\color\red{30}$
  • The maximum value of $n$ such that $(\color\green{3^1})^n$ divides $(48!)^2$ is $\Big\lfloor\frac{\color\green{22}\cdot2}{\color\green{1}}\Big\rfloor=\color\green{44}$

Therefore, the maximum value of $n$ such that $24$ divides $(48!)^2$ is $\min(\color\red{30},\color\green{44})=30$.

barak manos
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  • Thanks !! Got what I needed !! Just a doubt " Does $48! = 2^{46} . 3^{22}$ " ? Or there are some more numbers which when multiplied to $2^{46} . 3^{22}$ give $48!$ . – Jon Garrick Dec 29 '16 at 08:20
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    @Garrick: No, of course not. $48!=1\cdot2\cdot3\cdot\ldots\cdot48$. – barak manos Dec 29 '16 at 08:21
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    @Garrick: No, of course not!!! What other integers are between $1$ and $48$? – barak manos Dec 29 '16 at 08:24
  • ohh !! Right . My bad. I should be more vigilant . Thanks !! – Jon Garrick Dec 29 '16 at 08:26
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    @Garrick: You're welcome :) – barak manos Dec 29 '16 at 08:29
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    If the problem asked about $(50!)^2$ instead of $(48!)^2$: The maximum value of $n$ such that $(2^3)^n$ divides $(50)!$ is $\lfloor\frac{47}3\rfloor=15$ but the maximum value of $n$ such that $(2^3)^n$ divides $(50!)^2$ is $\lfloor\frac{94}3\rfloor=31\ne15\cdot2.$ – bof Dec 29 '16 at 08:39
  • @bof: You're right. I should slightly change the answer, since it fits $48$ incidentally... Done... – barak manos Dec 29 '16 at 08:42
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    @Garrick: Please note the slight modification in the answer (though for the specific case of $(48!)^2$ the bottom line remains the same), due to the comment above made by bof. – barak manos Dec 29 '16 at 08:46
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HINT Use the fact that $24=2^3\cdot3$ and apply Legendre's formula to compute the largest power of $2$ and $3$ in $48!$, i.e. $$p_2=\sum_{i=1}^{\infty} \left\lfloor \frac {48}{2^i} \right\rfloor $$ $$p_3=\sum_{i=1}^{\infty} \left\lfloor \frac {48}{3^i} \right\rfloor$$ For $2$ and $3$ respectively.

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Divide $48$ by $2$ repeatedly and add the quotient omitting any remainder,

$48/2=24, 24/2=12, 12/2=6, 6/2=3, 3/2=1$(with remainder as $1$). Therefore adding the quotients,i.e, $24+12+6+3+1=46$ $\Longrightarrow$ there are $46$ $2$s in $48!$ and $92$ $2$s in $(48!)^2$.

Similarly, divide 48 by 3 repeatedly and add the quotients omitting any remainder,

$48/3=16, 16/3=5$ (with remainder as $2$), $5/3=1$ (with remainder as $2$). Adding these quotients gives , $16+5+1=22$ $\Longrightarrow$ There are $22$ $3$s in $48!$ and hence $44$ $3$s in $(48!)^2$.

Now, the number of $24$s ($2^3\times3$)that can be made from $92$ $2$s and $44$ $3$s is $30$.

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    how is the number of $24$s that can be made from $46$ $2$s and $22$ $3$s equal to $5$, because a $24$ requires just $3$ $2$s and one $3$, isnt it? – vidyarthi Dec 29 '16 at 04:51
  • Thanks Garrick for pointing that out. I edited my answer. –  Dec 29 '16 at 05:07
  • @Vidyarthi : I am not talking about the number of 2s and 3s in 24 but the "number of 24s" that can be made using 46 twos and 22 threes. Sorry if it's not coming out that clear in the answer. –  Dec 29 '16 at 05:09
  • you just now edited it. It was actually $15$, but you wrote $5$, which was the cause of all confusion. – vidyarthi Dec 29 '16 at 05:26
  • Sorry for that, I hope it's clear now after the edit –  Dec 29 '16 at 05:37
  • @naveendankal Why multiply by 2 atlast ? – Jon Garrick Dec 29 '16 at 06:23
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    The question asks for (48!)$^2$ and not 48!. –  Dec 29 '16 at 06:27
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    The question asks for (48!)2 and not 48!. Say if you are required to find the number of 3s that divide (3!)2. Now, only one 3 divides 3! and a total of two 3s divide $(3!)^2$ and not $1^2$. Hope that makes it clear. –  Dec 29 '16 at 07:00
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    You got the right answer but I think you want to multiply by $2$ one step earlier. Supposing there were not $46$ but $47$ $2$s in $48!$ then there would still be $15$ $24$s in $48!$ but $31$ $24$s in $(48!)^2.$ – bof Dec 29 '16 at 07:05
  • Thanks a lot for pointing that out. I have made the edit in my answer. –  Dec 29 '16 at 16:09