Wikipedia:Reference desk/Archives/Mathematics/2015 July 8

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July 8

Puzzle

I am trying to solve a puzzle set by someone who has a tendency to set "mathsy" ones. It is

N51 11. M(12) + M(7) + M(6) - M(1)
W000 38. M(10) + M(8) +M(4) - M(2)
I think you will know if you have solved this puzzle but as a check, 
the sum of all 15 digits is 39.

The two expressions will result in numbers in the range 0 to 999. I am trying to come up with something that the M(x) could be. My first (and only so far) idea was Mersenne numbers, but M(12) is too big. Anyone got any suggestions for what function M(x) could be? Thanks -- SGBailey (talk) 13:33, 8 July 2015 (UTC)

Some background: The puzzle seems to be from [1], and is related to Geocaching. Knowing the geocaching connection, it's apparent that the N and W are latitude and longitude and from info on the site it's from somewhere in England. I thought the M's might refer to a the highway numbering system in Britain, but it turns out the M12 was proposed and never built, so unlikely. I doubt that the poser would make a clue that only a few mathies would understand, in fact I suspect that this is a case where knowing math is a hindrance rather than a help. --RDBury (talk) 14:16, 8 July 2015 (UTC)

I'm just trying to understand the question:

• 1) What are "11." and "38." ? Also part of the latitude and longitude ?

• 2) What "15 digits" add up to 39 ? Is that the digits in M(12) etc ?

• 3) Can M(x) be any function at all ?

• 4) Are we multiplying latitude and longitude by M(12) and M(10) ?

StuRat (talk) 14:23, 8 July 2015 (UTC)

The answer will be a long/lat such as N51 11.654, W000 38.122 where 5+1+1+1+6+5+4+0+0+0+3+8+1+2+2 = 39. The format used for each coord is N (or S or W or E) degrees minutes_to_3_dp. The puzzle answer(s) are therefore integer milliminutes. M(x) will be some function that was chosen by the puzzle setter, I've no idea what. He has done another one recently using (4,2) which turned out to be "select 2 from 4", another "mathsy" puzzle. No we aren't multiplying. -- SGBailey (talk) 14:35, 8 July 2015 (UTC)

Thanks. Let me try to restate that in more of a mathematical way:

M(12) + M(7) + M(6) - M(1) = a
M(10) + M(8) + M(4) - M(2) = b
0 ≥ a ≥ 999
0 ≥ b ≥ 999
count of digits in (M(12),M(7),M(6),M(1),M(10),M(8),M(4),M(2)) = 15 
total of digits in (M(12),M(7),M(6),M(1),M(10),M(8),M(4),M(2)) = 39  
(N51 11+a/1000, W000 38+b/1000) are geochache coords of a significant object

Is this all correct ? From the digit info, we can also infer that M(x) always produces an integer. StuRat (talk) 14:52, 8 July 2015 (UTC)

Yes M(x) always produces an integer. I don't understand your count of digits in ... =15. There are 15 digits in the final coordinate, represented by # in N ## ##.###, W ### ##.### . Of those 15 digits, 6 are the puzzle and 9 are given in the puzzle text. Then when you add those 15 digits together, you get 39. It is a function of the final answer, not the values of the individual M(x). -- SGBailey (talk) 21:26, 8 July 2015 (UTC)

As I understand it, there are 3 digits each in a and b, 4 digits in N51 11, and 5 digits in W000 38, which makes 15. so the count of digits in M(12), etc. could be anything. Also, I could be wrong so someone please check this, but after looking more carefully at the website it seems there is a picture you only get to see if you're logged on. So there may be an additional clue that we're not getting. --RDBury (talk) 16:23, 8 July 2015 (UTC)

PS. This [2] may be the missing clue. --RDBury (talk) 18:49, 8 July 2015 (UTC)

I had missed the photo - thanks for pointing it out. -- SGBailey (talk) 21:26, 8 July 2015 (UTC)

No photo here. :( My browser (IE11 @ Win7) says it displays 'only the safe contents', and when I decline from 'show everything' it lefts me with an empty window. --CiaPan (talk) 08:55, 9 July 2015 (UTC)

It's a pic of a piece of paper with a solved simultaneous equation on it. The specific equations were 5y - 4x = 8 and y + x = 7, with solutions x = 3, y = 4 (although the y solution is incorrectly listed as 4/5 on the paper, and marked wrong). So, the meaning of the pic is unclear to me. Is there meant to be a simultaneous equation in this problem, too ? StuRat (talk) 14:40, 9 July 2015 (UTC)

OK, I've struck my apparently incorrect statement of the problem with respect to the digits. Here's my current understand:

a(hundreds digit) + a(tens digit) + a(ones digit) + b(hundreds digit) + b(tens digit) + b(ones digit) + 5 + 1 + 1 + 1 + 0 + 0 + 0 + 3 + 8 = 39

a(hundreds digit) + a(tens digit) + a(ones digit) + b(hundreds digit) + b(tens digit) + b(ones digit)  = 20

Is that correct ? StuRat (talk) 22:37, 8 July 2015 (UTC)

Yes... -- SGBailey (talk) 08:42, 9 July 2015 (UTC)

To solve this, I think a program might help. Try out various functions for M(x). When it finds one that produces the total of 20 for the digits, and a and b are in the proper range, it should bring up those coords in a map, to see if there is anything of interest there. Hopefully that would be a small set, so the person looking at them could quickly figure out whether that's the desired solution. The tricky part would be finding the list of functions to try. StuRat (talk) 16:41, 11 July 2015 (UTC)

You are assuming M is a function. You are also assuming M is a simplistic function, such as M(x) = x+b for some value of b. Even if you use a simple function (like the x+b one), you get many answers. If M(x) = x+b and you loop through all values of b from 0 to 10000, you get many possibilities for the first and second three-digit values: 037,028 064,046 091,064 118,082 172,118 226,154 253,172 280,190 307,208 334,226 361,244 415,280 523,352 550,370 604,406 631,424 820,550 901,604. 199.15.144.250 (talk) 18:21, 11 July 2015 (UTC)

Didn't the question define M as a function ? The looping is a good start, but there's no reason for the loop to go that high. Remember that M(12) + M(7) + M(6) - M(1) is 999 or less, and so is M(10) + M(8) + M(4) - M(2). With that in mind, the C loop only needs to go up to about 500, where M(x) = x + C. StuRat (talk) 21:09, 11 July 2015 (UTC)

The clue you cannot see if you are not a Geocaching member is "One mark off". 199.15.144.250 (talk) 18:46, 11 July 2015 (UTC)

What does that mean ? StuRat (talk) 21:09, 11 July 2015 (UTC)

In case anyone is still looking for an answer to this: the homework in the picture had dropped a mark, which as the additional clue said, meant it was only "one mark off"... "mark off"... "Markov"... Markov number. Which gives you N51 11.732, W000 38.413. 129.234.186.11 (talk) 10:56, 14 July 2015 (UTC)

Thank you, -- SGBailey (talk) 14:32, 14 July 2015 (UTC)

OK, let's check it. Here's the Markov numbers: [3]. The first 12 are 1, 2, 5, 13, 29, 34, 89, 169, 194, 233, 433, 610. Let's plug those in:

a = M(12) + M(7) + M(6) - M(1) = 610 + 89 + 34 - 1 = 732
b = M(10) + M(8) + M(4) - M(2) = 233 + 169 + 13 - 2 = 413

Both a and b are between 0 and 999. Digits of 7, 3, 2, 4, 1, 3 add up to 20. The coords are:

N51 11.732, W000 38.413

Which seems to be Hartsmere Cottage, Hurtmore, Hurtmore Bottom, Godalming GU7 2RP, UK. Not sure what the significance of that location is (although you have to love the name "Hurtmore Bottom"). StuRat (talk) 14:50, 14 July 2015 (UTC)

There are quite few areas called Hurtmore in Surrey - there's probably some linguistic meaning to it. The actual "treasure" (tupperware box with a log book in it for signing) will be in the trees to the south of the path rather than in the cottage to the north (indeed there is a ROT13 hint on the page saying it is under a log new a yew tree). (SR, I fixed your 99 -> 999 above). SR - thanks for checking the checksum. And for what it is worth, there is a "geochecker" link on the page which confirms that this answer is indeed the correct one. ((( If anyone can't see the webpage we are talking about, you can get a free username from that website's main page that will allow access.)))-- SGBailey (talk) 12:30, 15 July 2015 (UTC)

Flight time

Hi, if my flight departs Birmingham, United Kingdom at 8:55 am (BST), and arrives in New York at 11:55 (EST), how long is my flight. If we subsequently depart New York at 2:55 pm and arrive in Orlando, Florida at 6:05 pm, what is the total time that I have spent flying. Could you also give me a formula to figure this out for myself - I think I was taught in school, but forgot. --Andrew 17:20, 8 July 2015 (UTC)

Google will convert time zones for you, so you can search "11:55 AM EDT in BST" and it will quickly tell you that 11:55 Eastern Daylight (Saving) Time (thus EDT, not EST) is 4:55 PM BST. So the first leg of your trip is 8 hours. New York and Orlando are in the same time zone, so you just have to add the 3:10 to your original 8 hours to find out your total airline time is 11 hours, 10 minutes. Of course this does not account for boarding, taxiing, etc. --LarryMac | Talk 19:48, 8 July 2015 (UTC)

Actually it does include taxiing. Airline schedules are based on gate-to-gate times, not takeoff-to-landing. See e.g. this. That's why, when you board the plane in Birmingham, you may be surprised to hear an announcement like "our flying time will be 7 hours 25 minutes"; that is a takeoff-to-landing time. --174.88.133.209 (talk) 21:29, 8 July 2015 (UTC)

Generalization of collinear and coplanar to higher dimensions

Collinearity and coplanarity can be generalized to higher dimensions. Just as two points are always collinear and three points are always coplanar, n points always lie in the same (n − 1)-dimensional space. GeoffreyT2000 (talk) 20:31, 8 July 2015 (UTC)

Is there a question? --174.88.133.209 (talk) 21:29, 8 July 2015 (UTC)

I suppose that if there is a question, it is whether the OP's statement it true or not. AndyTheGrump (talk) 21:36, 8 July 2015 (UTC)

Yes, it's true. Note that $S={\text{span}}\{v_{2}-v_{1},v_{3}-v_{1},\ldots ,v_{n}-v_{1}\}$ is an (n-1)-dimensional or smaller vector space, and $v_{1}+S$ contains all the n-vectors. Abecedare (talk) 21:48, 8 July 2015 (UTC)

On Win7 & MSIE, Abecedare'a maths expressions do not show. -- SGBailey (talk) 10:34, 9 July 2015 (UTC)

To repeat in plain-text: Note that S, defined to be the span of the vectors (v2-v1), (v3-v1),..., (vn-v1), is an (n-1)-dimensional or smaller vector space, and v1+S contains all the n-vectors.

Anyone know how to resolve the issue in general? Abecedare (talk) 14:52, 9 July 2015 (UTC)

Use a better browser? More seriously, Help:Displaying_a_formula has some info. You could try doing it in html, or see the section on forcing re-rendering. The <math> environment just generates regular .png images, the one you typed is here [4]. I have no idea why IE on Win7 is failing, I don't think that's a general problem, might have to do with SGB's setup. If they can't get an image from my last link, then it's almost surely a localized problem for them. SemanticMantis (talk) 16:48, 9 July 2015 (UTC)

It's worth noting that the article Affine space is relevant. Sławomir Biały (talk) 21:56, 8 July 2015 (UTC)

Another question might be what you call that general concept. StuRat (talk) 23:06, 8 July 2015 (UTC)

See flat (geometry) and hyperplane. Gandalf61 (talk) 14:09, 9 July 2015 (UTC)

Thanks for the links! I wasn't familiar with the first term, and used to use "hyperplanes" for the concept described by "flat" instead. Abecedare (talk) 14:52, 9 July 2015 (UTC)

Linear subspace could also be relevant. RJFJR (talk) 17:04, 9 July 2015 (UTC)