FACTORIAL NOTATION
- Thread starter r267747
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One way to consider the two questions in the first exercise is to think about prime factorizations for the even factors in the numerator. In other words, just how many factors of 2 are up there?
\(\displaystyle \frac{32!}{2^{16}}\)
32! is small enough to write it all out.
\(\displaystyle \frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11\cdot12\cdot13\cdot14\cdot15\cdot16\cdot17\cdot18\cdot19\cdot20\cdot21\cdot22\cdot23\cdot24\cdot25\cdot26\cdot27\cdot28\cdot29\cdot30\cdot31\cdot32}{2^{16}}\)
\(\displaystyle \frac{1\cdot2\cdot3\cdot(2\cdot2)\cdot5\cdot(2\cdot3)\cdot7\cdot(2\cdot2\cdot2)\cdot9\cdot(2\cdot5)\cdot11\cdot(2\cdot2\cdot3)\cdot13\cdot(2\cdot7)\cdot15\cdot16\cdot\hdots\cdot29\cdot(2\cdot3\cdot5)\cdot31\cdot(2\cdot2\cdot2\cdot2\cdot2)}{2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2}\)
There are obviously at least 16 factors of 2 in the numerator, to cancel with 16 factors of 2 in the denominator (2^16). So, yes, 2^16 will divide evenly.
See how it goes?
\(\displaystyle \frac{32!}{2^{16}}\)
32! is small enough to write it all out.
\(\displaystyle \frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11\cdot12\cdot13\cdot14\cdot15\cdot16\cdot17\cdot18\cdot19\cdot20\cdot21\cdot22\cdot23\cdot24\cdot25\cdot26\cdot27\cdot28\cdot29\cdot30\cdot31\cdot32}{2^{16}}\)
\(\displaystyle \frac{1\cdot2\cdot3\cdot(2\cdot2)\cdot5\cdot(2\cdot3)\cdot7\cdot(2\cdot2\cdot2)\cdot9\cdot(2\cdot5)\cdot11\cdot(2\cdot2\cdot3)\cdot13\cdot(2\cdot7)\cdot15\cdot16\cdot\hdots\cdot29\cdot(2\cdot3\cdot5)\cdot31\cdot(2\cdot2\cdot2\cdot2\cdot2)}{2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2}\)
There are obviously at least 16 factors of 2 in the numerator, to cancel with 16 factors of 2 in the denominator (2^16). So, yes, 2^16 will divide evenly.
See how it goes?
Hello, r267747!
\(\displaystyle 32!\text{ contains the factors: }\:1,2,3,4,\,\hdots,32\)
\(\displaystyle \text{Every }2^{nd}\text{ factor contains }2^2\text{ and each contributes a factor of 2: }\quad\frac{32}{2} \;=\;16\text{ factors of 2.}\)
\(\displaystyle \text{But every }4^{th}\text{ factor contains }2^2\text{ and each contributes an additional 2: }\quad\frac{32}{4}\,=\,8\text{ more factors of 2.}\)
\(\displaystyle \text{And every }8^{th}\text{ factor contains }2^3\text{ which contributes another 2: }\quad \frc{32}{8} \,=\,4\text{ more factors of 2.}\)
\(\displaystyle \text{And every }16^{th}\text{ factor contains }2^4\text{ which contributes another 2: }\quad \frac{32}{16} \,-\,2\text{ more factors of 2.}\)
\(\displaystyle \text{And every }32^{nd}\text{factor contains }2^5\text{ which contributes another 2: }\quad\frac{32}{32}\,=\,1\text{ more factor of 2.}\)
\(\displaystyle \text{Hence, }\,32!\text{ contains: }\:16 + 8 + 4 + 2 + 1 \,=\, 31\text{ factors of 2.}\)
. . \(\displaystyle \text{That is, }32!\text{ contains a factor of }2^{31}\)
\(\displaystyle \text{Therefore, it is certainly divisible by }2^{16}.\)
We want the permutations of the digits {1, 3, 5, 7}.
And there are: .\(\displaystyle 4! \,=\,24\) such four-digit numbers.
\(\displaystyle 32!\text{ contains the factors: }\:1,2,3,4,\,\hdots,32\)
\(\displaystyle \text{Every }2^{nd}\text{ factor contains }2^2\text{ and each contributes a factor of 2: }\quad\frac{32}{2} \;=\;16\text{ factors of 2.}\)
\(\displaystyle \text{But every }4^{th}\text{ factor contains }2^2\text{ and each contributes an additional 2: }\quad\frac{32}{4}\,=\,8\text{ more factors of 2.}\)
\(\displaystyle \text{And every }8^{th}\text{ factor contains }2^3\text{ which contributes another 2: }\quad \frc{32}{8} \,=\,4\text{ more factors of 2.}\)
\(\displaystyle \text{And every }16^{th}\text{ factor contains }2^4\text{ which contributes another 2: }\quad \frac{32}{16} \,-\,2\text{ more factors of 2.}\)
\(\displaystyle \text{And every }32^{nd}\text{factor contains }2^5\text{ which contributes another 2: }\quad\frac{32}{32}\,=\,1\text{ more factor of 2.}\)
\(\displaystyle \text{Hence, }\,32!\text{ contains: }\:16 + 8 + 4 + 2 + 1 \,=\, 31\text{ factors of 2.}\)
. . \(\displaystyle \text{That is, }32!\text{ contains a factor of }2^{31}\)
\(\displaystyle \text{Therefore, it is certainly divisible by }2^{16}.\)
We want the permutations of the digits {1, 3, 5, 7}.
And there are: .\(\displaystyle 4! \,=\,24\) such four-digit numbers.