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The largest non-negative integer $k$ such that $24^{\mathrm{k}}$ divides $13 !$ is.
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The correct answer is:
3
$24^{k} \rightarrow\left(2^{3} \times 3\right)^{k}$
Exponent of 2 in 13! $\left[\frac{13}{2}\right]+\left[\frac{13}{2^{2}}\right]+\left[\frac{13}{2^{3}}\right]=10$
Exponent of 3 in $13 !$ $\left[\frac{13}{3}\right]+\left[\frac{13}{3^{2}}\right]=5$
So $\left(2^{3} \times 3\right)^{3}$ So $\mathrm{K}=3$
Exponent of 2 in 13! $\left[\frac{13}{2}\right]+\left[\frac{13}{2^{2}}\right]+\left[\frac{13}{2^{3}}\right]=10$
Exponent of 3 in $13 !$ $\left[\frac{13}{3}\right]+\left[\frac{13}{3^{2}}\right]=5$
So $\left(2^{3} \times 3\right)^{3}$ So $\mathrm{K}=3$
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