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Math Challenge II-A Combinatorics Week 4 Questions P3

 
 
JinBryant的头像
Math Challenge II-A Combinatorics Week 4 Questions P3
JinBryant - 2017年06月5日 Monday 23:27
 

I didn't understand the solution to P3. 81000 = 2^3 * 3^4 * 5^3. I didn't know why the solution said that 81000 = 2^4 * 3^4 * 5^4. I also don't understand why we can use a+b+c = 4 and why we should use it three times.

 
ReynosoDavid的头像
Re: Math Challenge II-A Combinatorics Week 4 Questions P3
ReynosoDavid - 2017年06月6日 Tuesday 11:09
 
The factoring of $81000$ is indeed incorrect, it should be $81000$ as $2^3\cdot3^4\cdot5^3$.

Now, each possible factor of $81000$ is of the form $2^x\cdot 3^y\cdot 5^z$. When we multiply the three factors we want, we need that the three exponents of the $2$s and the $5$s add up to $3$ and that the exponents of the $3$s add up to $4$. That is, we need to find how many non-negative solutions are to $a+b+c=3$, and how many non-negative solutions are to $a+b+c=4$, which we can do with stars and bars. Thus the number of possible ways of factoring $81000$ as three factors will be $\binom{3+3-1}{3}\cdot\binom{4+3-1}{4}\cdot\binom{3+3-1}{3}=1500$.


(The handout should now be updated on Edurila, thank you for letting us know!)

JinBryant的头像
Re: Math Challenge II-A Combinatorics Week 4 Questions P3
JinBryant - 2017年06月6日 Tuesday 16:39
 

Okay. Thank you!