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

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

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 $\min(\color\red{30},\color\green{44})=30$.


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.


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 $30$.

Tags:

Discrete Mathematics

Number Theory

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