Find the ones digit of the largest power of 2 that divides into (2^4)!.

To find the largest power of 2 that divides into (2⁴)!:

First:  simplify (2⁴)!   --->   (2⁴)!  =  (16)!

16! = 16 x 15 x 14 x ... x 2 x 1

But, if we want to find a number that divides into 16!, we don't really care about the odd numbers in  16 x 15 x ... x 2 x 1.

All we need to look at are the even factors:  16 x 14 x 12 x 10 x 8 x 4 x 2.

16  = 2⁴

14 = 2 x 7

12 = 2² x 3

10 = 2 x 5

8 = 2³

6 = 2 x 3

4 = 2²

2 = 2

So, the total number of factors is 4 + 1 + 2 + 1 + 3 + 1 + 2 + 1  =  15     <---   Answer

As a check:  16! / 2¹⁵  =  638 512 875  (no further factors of 2)

Find the ones digit of the largest power of 2 that divides into (2^4)!.

[...] integer Part

power of 2:

\(\begin{array}{|rcll|} \hline && \left[ \frac{2^4}{2^1} \right] + \left[ \frac{2^4}{2^2} \right] + \left[ \frac{2^4}{2^3} \right] + \left[ \frac{2^4}{2^4} \right] \\\\ &=& 2^3 + 2^2 + 2^1 + 1 \\\\ &=& 8+4+2+1 \\\\ &=& 15 \\\\ && 2^4! = 2^{15} \cdot \dots \\\\ \hline \end{array} \)

power of 3:

\(\begin{array}{|rcll|} \hline && \left[ \frac{16}{3^1} \right] + \left[ \frac{16}{3^2} \right] \\\\ &=& 5 + 1 \\\\ &=& 6 \\\\ && 2^4! = 2^{15} \cdot 3^6 \dots \\\\ \hline \end{array}\)

power of 5:

\(\begin{array}{|rcll|} \hline && \left[ \frac{16}{5^1} \right] \\\\ &=& 3 \\\\ && 2^4! = 2^{15} \cdot 3^6 \cdot 5^3 \dots \\\\ \hline \end{array} \)

power of 7:

\(\begin{array}{|rcll|} \hline && \left[ \frac{16}{7^1} \right] \\\\ &=& 2 \\\\ && 2^4! = 2^{15} \cdot 3^6 \cdot 5^3 \cdot 7^2 \dots \\\\ \hline \end{array}\)

power of 11:

\(\begin{array}{|rcll|} \hline && \left[ \frac{16}{11^1} \right] \\\\ &=& 1 \\\\ && 2^4! = 2^{15} \cdot 3^6 \cdot 5^3 \cdot 7^2 \cdot 11^1 \dots \\\\ \hline \end{array}\)

power of 13:

\(\begin{array}{|rcll|} \hline && \left[ \frac{16}{13^1} \right] \\\\ &=& 1 \\\\ && 2^4! = 2^{15} \cdot 3^6 \cdot 5^3 \cdot 7^2 \cdot 11^1 \cdot 13^1 \\\\ \hline \end{array} \)

\(16! = 2^{15} \cdot 3^6 \cdot 5^3 \cdot 7^2 \cdot 11 \cdot 13\)