Discussion Forum

Weekly Brain Potion Request and Idea Thread

 
 
LensmireJohn的头像
Weekly Brain Potion Request and Idea Thread
LensmireJohn 发表于 2018年10月12日 Friday 12:51
 

Hello all!

We had two main goals when coming up with the idea of the "Weekly Brain Potion" last year. One, we wanted to spark discussion among students while they solved interesting or thought provoking questions/riddles. Two, we hoped to expose students to new concepts and ideas or new applications of mathematical ideas.

Especially with the second goal, your feedback as users is invaluable! We're dedicating this thread as a discussion space where you can share your ideas. All ideas are helpful, whether your idea is just a new topic, concept, or application that you want to challenge us to address in a future brain potion or your idea it is a riddle or puzzle you've already seen that you think could be expanded on/discussed more.

We look forward to hearing your thoughts below!

 
ChengAlan的头像
Re: Daily Magic Spells
ChengAlan 发表于 2018年12月17日 Monday 18:25
 

Hello. My name is Alan Cheng, and I may have caught a few mistakes in the "Daily Magic Spells" section. The first is from December 8, 2018. I found that the probability of selecting a number that would not change the median is 1/10, because only one number out of the first ten integers does not change the median. Could someone tell me why the probability was 6/10? Also, for  December 10, 2018, the wording is slightly incorrect. It should be that x and y are POSITIVE integers, otherwise the largest sum is infinite. Even though this does not have anything to do with Weekly Brain Potions, It was the closest I could find to a forum. Thank you!

ReynosoDavid的头像
Re: Daily Magic Spells
ReynosoDavid 发表于 2018年12月19日 Wednesday 11:58
 

Hey Alan, thanks for your feedback! If you need to ask a question about anything that goes on on the ZIML site (Daily Magic Spells, Monthly Comptetition, etc.) feel free to start a new topic for discussion on the forum :)

Note on Dec 10 the problem is asking for the probability that the number will "not increase" the median, so it is ok if it stays the same or if it decreases, hence the $6$ possible values.

On the problem on Dec 8, since we know $x$ and $y$ satisfy a linear equation, we can write $x^2-y^2$ in terms of only $x$, which yields the quadratic function $-0.096x^2 + 31.92x -129.96$. This is a parabola that opens downwards, and thus has a global maximum. That is, the sum cannot grow indefinitely. You are however, correct, that the problem should ask for the values of $x$ and $y$ to be both positive, otherwise the presented solution is incorrect. Note the parabola $-0.096x^2 + 31.92x -129.96$ has a global maximum when $x = 16.625$, so the maximum value of $x^2 - y^2$ will occur when $x$ is as close as possible to $16.625$. All $x\equiv 1 \pmod{5}$ yield an integer solution to the equation, so $x = 16$ is the closest one. When $x = 16$, $y = -11$, so $16^2 - (-11)^2 = 135$ is the maximum possible value for $x$ and $y$ integers, and $6^2 - 3^2 = 27$ is the maximum value for $x$ and $y$ positive integers.