Comments for The Math Factor Podcast

Comment on Follow Up: The Harmonic Series by Susannah

Susannah — Fri, 08 Aug 2014 12:52:06 +0000

I’ve come across this while trying to discover the rules for constucting perspective-drawing of a series of columns of equal height equally-spaced. Having figured that the ratio of the apparent height each to the front-most one forms the series (proportional to) 1/2, 1/3, 1/4, 1/5 etc, I thought I could maybe draw an interesting spiral in which consecutive loops were separated by intervals growing in this series. No dice! If you start from the outside loop, stepping in by 1 unit in 360degrees, then draw the next loop stepping in by 1/2 unit, then by 1/3 etc, I think you can never locate a ‘centre’, because the radius = 1/2 + 1/3 + 1/4 etc. Am I right? Or am I missing something…

Comment on HR. CardColm by Pratik Poddar

Pratik Poddar — Mon, 16 Jun 2014 17:15:24 +0000

As posted in comments on my blog: http://www.cseblog.com/2013/07/self-referential-problem-from-what.html

For general N>6, the following solution works.

The number of 1?s in this paragraph is N-3; the number of 2?s is 3; the number of 3?s is 2; .. the number of N-3?s is 2. All other blanks are 1.

This solution works as long as N-3 > 3, i.e., N > 6.

Thought process
Again, the sum of the blanks has to be 2N. That implies the average in a blank is 2. This must mean most values are small.

Now, most blanks are small imply the majority of the values are 1 or 2. Most blanks cannot have 2 because then we won’t be able to satisfy all the statements ‘The numbers of K’s in the paragraph are 2’.

So, let us assume that most blanks are 1 and try possible values for the first blank. Suppose we fill the first blank with N-p. Then the (N-p)th blank will have value at least 2. A larger value will again lead to all constraints not being satisfied. So, the (N-p)th blank is 2.

Now, the second blank is at least 2. But it cannot be 2 (because then there would be three 2’s). So, the second blank is greater than equal to 3. As it turns out, it cannot be greater. So, this blank is filled with 3.

Now, the third blank can be filled with 2 and all others with 1. And the value of p can be determined by the fact that the sum is 2N. Comes out to be 3.

I agree the proof is not rigorous about why no other solution works.

Comment on A Quick Puzzle From OSCON by Paul H

Paul H — Thu, 08 May 2014 21:37:33 +0000

Hi, I just discovered your podcast today. I listened to several episodes and really enjoyed them. But it looks like you haven’t had any new episodes since 2012. Do you have any plans to produce more episodes?

Comment on GM. What’s the Big Deal Anyway? by Shawn

Shawn — Mon, 21 Apr 2014 01:42:40 +0000

Maybe it only works for digits that are not significant figures. The significant figures of a number are those digits that carry meaning contributing to its precision. This includes all digits except:

All leading zeros;
Trailing zeros when they are merely placeholders to indicate the scale of the number
Spurious digits introduced, for example, by calculations carried out to greater precision than that of the original data, or measurements reported to a greater precision than the equipment supports.

e.g., For A.D. 2009, all digits are significant, but for May 05, the 0 is not significant.

Comment on DH. Ice Cream Cake by Shawn

Shawn — Mon, 31 Mar 2014 19:17:01 +0000

I should probably be practicing differential equations before my class tomorrow instead of analyzing a puzzle and posting my results on a post that hasn’t had comments in over six years, but I don’t often do what I should. I haven’t read the solution in the book, but I have thought about this problem for many hours. Here goes: Let us imagine the ice cream cake is a right circular cylinder. If the volume of the cake is the volume bounded from below by the plane z = 0, from above by the plane z = 1, and constrained by the inequality x^2 + y^2 <= 1, we see that its projection on the x,y plane is the unit circle. Let us make our first cut of the ice cream cake such that our angle x (which I will refer to as ? to avoid confusion with the Cartesian coordinates) is an angle in standard position (i.e., the initial side will start at the line x=y=0 and go through the point (1,0,0) and the terminal side will be ? radians counterclockwise on the x,y plane). There is not a straightforward interpretation for what to do for |?| > 2pi nor for imaginary values of ?. Also, negative ? values can be simply thought of as its negation in the other direction, which for the purposes of this problem will yield symmetric and similar results. Therefore, I shall only concern myself with angles weakly between 0 and 2pi. For ? = 0, the problem is trivial. The terminal side and initial side of the angle will be the same, so there will be no cake to flip. After any number of flips, the icing will be (“back”) on top of the cake. For an angle of 2pi, the terminal side and initial side will again be the same, but this time the entire cake is the slice. After any even number of flips, the icing will be all on the top. As Peter Winkler said, an integer divisor of 2pi will also result in all the icing returning to the top, as during the second revolution, the slices made will be the exact slices as the previous revolution, and thus all slices will be flipped. The analysis gets slightly more complicated, however, for other values of ?. One thing that I have found that is important to keep in mind, and which I imagine Chaim was referring to above, is that [spoiler] when you flip a slice that is part icing on the bottom and part icing on the top, not only will the icing change from top to bottom and bottom to top, but also because the entire slice is rotated, the icing is also reflected across the line given by the terminal side of the angle, so to speak. For example, if you flip a slice whose top is 2/3 icing in a contiguous region contacting the terminal angle, the result will a slice whose top has 2/3 not icing, but touching the initial side of the angle. If you don’t understand what we mean, try drawing a slice of cake with icing the way we described and actually flip it.[/spoiler]

Comment on Follow Up: Escaping the Beast by jay ha

jay ha — Mon, 24 Mar 2014 21:55:38 +0000

what about this variation?

An escaped prisoner finds himself in the middle of a square swimming pool. The guard that is chasing him is at one of the corners of the pool. The guard can run faster than the prisoner can swim. The prisoner can run faster than the guard can run. The guard does not swim. Which direction should the prisoner swim in in order to maximize the likelihood that he will get away?

Comment on Follow Up: The Harmonic Series by david

david — Thu, 21 Feb 2013 11:44:35 +0000

ok…but what is the practical meaning of harmonic numbers?
are they numerical representations of wavelength and frequency?
or are they resonant frequency math expressions?

for example what does 1.3333 mean in terms of the real world….

how does pi relate to harmonic numbers?

sincerely,

math lover

Comment on HN. Barbette by Marcos

Marcos — Sat, 16 Feb 2013 02:44:38 +0000

I thought of a way to “ruin” your bet: instead of writting numbers directly, I write numerical expressions so difficult to compute or compare that the choice you make is effectively random. Later, I show you which number is the highest :)

Comment on Morris: A Day at the Races by sohbet

sohbet — Wed, 06 Feb 2013 15:38:46 +0000

three fastest from the first race compete in the second. The three fastest from the second compete in the third and so on, in 11 races you have the three fastest and the 22 other horses cycle in behind

Comment on Yoak: Will A Real Gold Coin Please Stand Up? by Parallel Minded

Parallel Minded — Tue, 29 Jan 2013 18:34:08 +0000

Stephen: Ah, nice! I hadn’t noticed that [spoiler]
the 3rd and 4th rounds work fine even for groups of a single coin, because you can group the comparisons like this:

Round 3:
1 3
2 4
5 7
6 8
…

Round 4:
3 5
4 6
7 9
8 10
…
[/spoiler]

The last question you ask is interesting, it’s like sorting algorithms where all the comparisons need to be specified ahead of time: Sorting Networks.

But I think it has a rather different answer… [spoiler]
You need to compare every possible pair of coins.

If there are any two coins A and B which are not compared with each other, then suppose all other coins (call them “class C coins”) are equal to each other but unequal to A or B. In this case you cannot know whether A and B are the same or not.
[/spoiler]
Would it help if you are told there is at least one of each type of coin?