Wikipedia:Reference desk/Archives/Mathematics/2008 September 25

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

September 25

MATH RIDDLE THAT I'VE BEEN TRYING TO FIGURE OUT FOREVER

I am a two dimensional figure. I have two pairs of parallel sides. None of my angles is a right angle.. All of my sides are the same length. What am a I_______________________________________ 75.34.60.50 (talk) 04:40, 25 September 2008 (UTC)ANGEL V.75.34.60.50 (talk) 04:40, 25 September 2008 (UTC)

Rhombus comes to mind. Digger3000 (talk) 04:59, 25 September 2008 (UTC)

A no longer puzzled rhombus? :) A bit more puzzling would be if you only had one pair of parallel sides if you can figure what that looks like but I don't think it has any special name. Dmcq (talk) 08:14, 25 September 2008 (UTC) Oh stupid me, they would be all on one line if all the sides are equal in length. I meant a rectangle arrangement of lengths but not a rectangle. Dmcq (talk) 08:20, 25 September 2008 (UTC)

Two pairs of parallel sides, not all sides the same length, is a parallelogram. One pair of parallel sides, not all sides the same length, is a trapezium. Darkhorse06 (talk) 09:49, 25 September 2008 (UTC)

Or in North America a trapezoid. Algebraist 10:15, 25 September 2008 (UTC)

Obviously a Quadrilateral ;-) hydnjo talk 11:27, 25 September 2008 (UTC)

...but I see nothing that says the figure has only four sides ;-) Gandalf61 (talk) 13:46, 25 September 2008 (UTC)

I'm struggling to think of a figure that has exactly two pairs of parallel sides and all sides equal but isn't a quadrilateral. (I know, I added the "exactly" bit.) Can this figure exist? Zain Ebrahim (talk) 14:05, 25 September 2008 (UTC)

"Exactly" two pair? Probably not. "At least" two pair? Quite. --LarryMac | Talk 14:15, 25 September 2008 (UTC)

Yeah, I was just asking out of curiosity. Zain Ebrahim (talk) 14:36, 25 September 2008 (UTC)

Take a regular hexagon with unit sides. Construct two adjacent isosceles triangles on one side, each of which has two sides with unit length, but with different bases that sum to 1 - say one has base length 1/3 and the other has base length 2/3. I think you now have an equilateral convex concave nonagon with exactly two pairs of parallel sides. Gandalf61 (talk) 14:18, 25 September 2008 (UTC)

Thanks, that works. Zain Ebrahim (talk) 14:36, 25 September 2008 (UTC)

If I've followed your construction correctly, it's not convex. Take a point within one triangle and a point within the other, the straight line between them goes outside the shape (that's the definition of convex, yes?). --Tango (talk) 14:48, 25 September 2008 (UTC)

Convex and concave polygons agrees with you. Zain Ebrahim (talk) 14:54, 25 September 2008 (UTC)

Yes, I meant to say concave. Fixed above. Gandalf61 (talk) 14:57, 25 September 2008 (UTC)

Gandalf, you do think outside the box however many sides that box has! -hydnjo talk 22:00, 25 September 2008 (UTC)

If you want it convex, that can be done. Take a regular decagon. Perturb one side, fixing all vertices not endpoints of that segment. This side and the sides next to it will no longer be parallel to the sides opposite them, but also won't be parallel to anything else in the figure. The remaining four sides will still form two pairs of parallel segments. Black Carrot (talk) 13:35, 26 September 2008 (UTC)

Integration in Matlab

Please help me do the following in MATLAB

Integrate(lower:416; upper:inf) 1250(x-416)D'(x)dx where D'(x)~(400, 900) i.e. D'(x) is normally distributed.

Please show me the commands I give.

I am not sure how to use appropriate symbols here. Hope you understand the problem.--131.204.11.21 (talk) 18:41, 25 September 2008 (UTC)

One possibility would be using some function performing numerical integration (for example, "quad" - input parameters would be handle of the function you want to integrate and lower and upper limits). Then you might have to guess what upper limit would be large enough to give a reasonable result and small enough to give a result in reasonable time (however, as far, as I understand, you could actually use "Inf" as a limit using function "quadgk" instead of "quad"). And the ways to display a formula in Wikipedia are described in Help:Displaying a formula. --Martynas Patasius (talk) 00:34, 26 September 2008 (UTC)

8,101,265,822,784

"To divide the number 8,101,265,822,784 by 8, all we need do is to transfer the 8 from beginning to end. Can you find a number beginning with 7 that can be divided by 7 in the same simple manner?" How would you solve this without using a using a computer program? Nadando (talk) 21:34, 25 September 2008 (UTC)

(1) 7101449275362318840579 ÷ 7 = 1014492753623188405797

(2) No clue. -hydnjo talk 22:07, 25 September 2008 (UTC)

Personally? I wouldn't. Looking at the sizes of the numbers in question, it doesn't seem that it could be easily found by hand, but let's see: What you need to do is find a number, x, such that ${\frac {x+7\cdot 10^{\lfloor \log _{10}x\rfloor +1}}{7}}=10x+7$. That reduces to $x={\frac {7\cdot 10^{\lfloor \log _{10}x\rfloor +1}-49}{69}}$. I doubt that can be solved analytically, I certainly can't do it. --Tango (talk) 22:27, 25 September 2008 (UTC)

(ec)It's not that hard. We require n a natural number, and x a natural number with at most n digits, such that (7*10ⁿ+x)/7 = 10x + 7, i.e. 7(10ⁿ - 7) = 69x. So we need an n such that (10ⁿ - 7) is divisible by 69. It's always divisible by 3, so we only have to worry about divisibility by 23. A quick pencil-and-paper calculation of the powers of 10 mod 23 reveals that we need n to be congruent to 21 mod 23 22. It's clear that each such n gives a unique x satisfying the given relation, and that x is always of at most n digits, as required. Calculating the x corresponding to the smallest possible n (i.e. 21) is just a matter of long division (and multiplication). Algebraist 22:32, 25 September 2008 (UTC)

To Tango: we don't actually need n to be log₁₀ x + 1, since we can pad x out with leading zeros if we want to. Algebraist 22:33, 25 September 2008 (UTC)

The calculating took me about 10 minutes, but still faster than learning to program. Algebraist 22:43, 25 September 2008 (UTC)

Youse guys are amazing, I had to look it up :( hydnjo talk 22:46, 25 September 2008 (UTC)

I was just about to ask how you'd found that, since I worked out the program I'd written was going to take years to find that answer. Admittedly, I'd written it in php which isn't particularly fast, but a better language wasn't going to cut it down to 30 minutes! Doing it a reasonable amount of time would have required a far clever algorithm than mine. Of course, Algebraist makes all that unnecessary by just being a genius! (Incidentally, I had worked out the padding leading zeros thing just before I read his comment, so there! ;)) --Tango (talk) 22:51, 25 September 2008 (UTC)

Out of curiosity, I used Algebraist's method (and a computer!) and found that the smallest one is, indeed, the one Hydnjo found. However, I'm finding that the 2nd smallest isn't an integer... I calculated ${\frac {7\cdot (10^{44}-7)}{69}}$, what blindingly obvious thing am I doing wrong? --Tango (talk) 23:00, 25 September 2008 (UTC)

You believed what I said (always a mistake!). I of course meant that n has to be congruent to 21 mod 22. Algebraist 23:06, 25 September 2008 (UTC)

Exercise for the reader with a better calculating program than Google calc (and too much time on their hands): if my calculations of the powers of 10 modulo various primes are correct (and they might well not be), this trick is possible with 7 replaced by any (non-zero!) digit. Algebraist 23:10, 25 September 2008 (UTC)

I can do it for 1, just take n=0! ;) (Any sequence of 1's will, of course, do, which satisfies my conjecture below. Proof by trivial example, that'll do for me!) --Tango (talk) 23:24, 25 September 2008 (UTC)

Ah, that'd do it! Ok, then, the next number is 71014492753623188405797101449275362318840579, which is the last number repeated twice - I didn't expect that, do you? I guess it makes a certain amount of sense, although I haven't worked it all through and proven it. --Tango (talk) 23:24, 25 September 2008 (UTC)

I didn't expect that, no, but now you mention it, it is obvious: just consider what happens when you divide the doubled-up number by 7 using long division. Algebraist 23:31, 25 September 2008 (UTC)

Tango, I looked it up (cheated), from here choose this. hydnjo talk 23:14, 25 September 2008 (UTC)

You would have been better off with the first result ([1]), which lists them for all the digits! Algebraist, I have a much better program than Google calc, I have Google! ;) --Tango (talk) 23:26, 25 September 2008 (UTC)

Yes, of course :( how silly of me to leave that part out ;) hydnjo talk 23:37, 25 September 2008 (UTC)

If using a computer program is cheating, using Sloane's is definitely cheating. It knows all, after all. Algebraist 23:31, 25 September 2008 (UTC)

The OP asked: "Can you find a number beginning with 7 that can be divided by 7 in the same simple manner..." Now youse guys are debating the method! hydnjo talk 23:41, 25 September 2008 (UTC)

It's a trick I often use - you work out the method, write it all out (or, get your local Ref Desk genius to do it for you! ;)) and then look up the answer and write that down at the end. It's much easier than actually calculating things! --Tango (talk) 23:49, 25 September 2008 (UTC)

Weaklings! Does no-one learn how to calculate any more? (waves piece of paper covered in numbers, mutters about country going to the dogs) Algebraist 23:52, 25 September 2008 (UTC)

I can calculate it, but it's quicker not to. I have similar pieces of paper scattered all over my room from previous exploits. The nearest one to hand (it somehow made its way under my keyboard) has some squiggles that look like something to do with SHM on most of it and a whole load of polynomials in the corner. I think they are both from previous ref desk questions! The country has gone to the dogs, though - I had to teach a 2nd year computer science student how to convert between grams and kilograms once, that was a low point... --00:11, 26 September 2008 (UTC)

It all started with the HP35 (and other calculators) being allowed into the classroom. When I ask my grandkid the sum of 17 and 19 or the product of 7 times 13 I get a blank stare. I'm very old fashioned about this but my grandkid ought to have some concept of size if not the exact answer but he doesn't. That's sad. Some grasp of numerical quantities should be natural, not mysterious. So, I send him flash cards (remember those?) so that he'll have some concept of quantities and size (if he doesn't just toss them away). I think that I'm starting to soapbox so... hydnjo talk 00:15, 26 September 2008 (UTC)

Flashcards? They're a means for rote memorisation, they don't actually teach anything... --Tango (talk) 00:20, 26 September 2008 (UTC)

Wrong! Size is important. Not trying to teach math here, just basic concepts of scale. 10 * 10 is a lot different than 10 + 10. I'm amazed by the lack of numerical scale comprehension in today's (my grandkid's) natural understanding. Yeah, flashcards don't teach math, they just expand the field a bit. But then I've already said about the soapboxing. hydnjo talk 00:33, 26 September 2008 (UTC)

If a kid doesn't get exposure to how numbers "connect" then that kid is less likely to be interested in exploring this arena. I don't mean to put any more meaning than that on this subject - early curiosity does develop. My Gilbert chemistry set certainly did not make me a chemist but it certainly piqued my interest in chemistry and other scientific endeavors. Devoid of those curious opportunities the young mind sometimes looses those opportunities. 8,101,265,822,784 is interesting, especially to a young mind. hydnjo talk 00:57, 26 September 2008 (UTC)

It's interesting to us slightly older minds too, judging by the length of this section! (Even ignoring the off topic small ramblings!) --Tango (talk) 02:04, 26 September 2008 (UTC)

Yeah, even when I promised to stop soap-boxing. I guess sometimes the passion takes over. I suppose that's why there are discussion forums ;) hydnjo talk 02:17, 26 September 2008 (UTC)

Exdenting for the curious:

210526315789473684 ÷ 2 = 105263157894736842

3103448275862068965517241379 ÷ 3 = 1034482758620689655172413793

410256 ÷ 4 = 102564

510204081632653061224489795918367346938775 ÷ 5 = 10204081632653061224489795918367346938775

6101694915254237288135593220338983050847457627118644067796 ÷ 6 = 1016949152542372881355932203389830508474576271186440677966

7101449275362318840579 ÷ 7 = 1014492753623188405797 (The original question.)

8101265822784 ÷ 8 = 1012658227848 (The original proposition.)

91011235955056179775280898876404494382022471 ÷ 9 = 10112359550561797752808988764044943820224719

hydnjo talk 03:13, 26 September 2008 (UTC)

I'm not a math guy, so I did this in a rather juvenile manner. I don't think it is the same method explained above. I assume that this is asking (7|X)/7=X|7 where the | is concatenation. If I assume Y = 10^{number of digits in X}, I get (7Y+X)/7=10X+7. That reduces easily to X=(69Y-49)/69. I know Y is 10 or 100 or 1000... So, I know that 69Y-49 is something like 6999...99951. So, X=6999...99951/69. I just do long division and keep using 9's until I get to a point that ending with 51 will be a perfect integer. It is rather easy to do with a pencil and a calculator. -- kainaw™ 05:48, 26 September 2008 (UTC)

This can of course be extended above 9 by moving 2 digits to the end. EG 10_0100 / 10 = 0100_10 (The _ is a separator of no nummeric significance). Thus we have (10Y+x)/10 = 100X+10. Presumably similar can be done in other bases as well. -- SGBailey (talk) 08:42, 26 September 2008 (UTC)

Not sure if anyone has already mentioned this, but we have:

${\frac {2}{19}}=.105263157894736842...$

${\frac {3}{29}}=.1034482758620689655172413793...$

...

${\frac {7}{69}}=.1014492753623188405797...$

${\frac {8}{79}}=.1012658227848...$

etc. because

$n+{\frac {n}{10n-1}}={\frac {10n^{2}}{10n-1}}=10n\left({\frac {n}{10n-1}}\right)$

so, as Kainaw says, any of these problems can be solved by solved in minutes with pencil and paper and long division (although I cheated and used a spreadsheet). Gandalf61 (talk) 09:51, 26 September 2008 (UTC)