November 2010 - mathfun@mailman.xmission.com

[math-fun] muffin problem -- clusters and bounds
by Huddleston, Scott 01 Dec '17

01 Dec '17

I enclose examples of some of the bounds in my 28 Feb. post, and a derivation of the most involved example. S(12/7) = 3/7 = r/4 = (3-r)/3 Muffins: 1/7 * { [3,4]*12 } Persons: 1/7 * { [4*3]*4, [3*4]*3 } S(7/4) = 5/12 = (3-r)/3 = (2r-1)/6 Muffins: 1/12 * { [5,7]*6, [6*2] } Persons: 1/12 * { [5*3,6]*2, [7*3]*2 } S(40/23) = 19/46 = (2r-1)/6 = (12-5r)/8 Muffins: 1/46 * { [19,27]*24, [20,26]*12, [23*2]*4 } Persons: 1/46 * { [19*3,23]*8, [20*4]*3, [27*2,26]*12 } S(228/131) = 54/131 = (12-5r)/8 = (7r-6)/15 Muffins: 1/131 * { [54,77]*120, [55,76]*24, [57,74]*60, [65,66]*24 } Persons: 1/131 * { [54*3,66]*24, [54*2,55,65]*24, [57*4]*15, [77*2,74]*60, [76*3]*8 } S(449/258) = 319/774 = (7r-6)/15 = (73-32r)/42 Muffins: 1/774 * { [319,455]*210, [320,454]*30, [325,449]*42, [336,438]*30, [337,437]*90, [384,398]*42, [387*2]*5 } Persons: 1/774 * { [319*3,390]*42, [319*2,325,384]*42, [320*3,387]*10, [336,337*3]*30, [455*2,437]*90, [455,454,438]*30, [449*3]*14 } Here's a derivation of 449/258 example. 449 muffins -> 898 pieces, used in 124 4-parts with 496 pieces and 134 3-parts with 402 pieces. We split muffins between a 3-part and a 4-part when possible, but 4-parts contain 94 more pieces than 3-parts. We divide these 94 pieces (47 muffins) as evenly as possible among the 124 4-parts. 47 pairs of 4-parts contain both halves of a muffin, which I'll call an A^1 cluster and write A^1 = [a1*3,m]*2 where m denotes the two muffin halves and the other pieces are denoted a1, indicating a set of pieces of possibly different sizes that occure in A^1 clusters. 34 4-parts contain only muffin pieces whose mates appear in 3-parts. These I call A^0 clusters and write A^0 = [a0*4] The 449/258 example uses 30 A^0 clusters ([a0*4])*30 (120 a0 pieces, 30 4-parts) 47 A^1 clusters ([a1*3,m]*2)*47 (282 a1 pieces, 94 4-parts) Note that the A^1 cluster by itself sets an upper bound on S -- the cluster is 2 persons (size 2r), it cntains 1 complete muffin (size 1, pieces m*2), and it has 6 unmatched muffin pieces (a1*6), giving S(449/258) <= avg(a1) = (2r-1)/6. A^0 clusters also bound S by S(449/258) <= avg(a0) = r/4, but the A^1 cluster bound is tighter. Note that this (2r-1)/6 bound holds for examples 7/4 and 40/23. The muffin mates of a0 pieces (resp. a1 pieces) are z0 pieces (resp. z1 pieces). We distribute z0 and z1 pieces as evenly as possible across the 134 3-parts (402 pieces). With 282 z1 pieces, at least 14 3-parts must consist of only z1 pieces. These are Z^0 clusters, Z^0 = [z1*3] = [z10*3] where I refine piece labels so that z1 pieces appearing in Z^0 clusters are labelled z10. I create general Z^k clusters by finding minimal sets of persons that contain complete a0/z0 muffins. Maximally even distribution of the 120 z0 pieces places one per 3-part. Then a Z^1 cluster, closed under a0/z0 muffins, is Z^1 = [a0*4],[z0,z1*2]*4 = A^0,[z0,z11*2]*4 where z1 pieces appearing in Z^1 clusters are relabelled z11. Z^0 clusters bound S by S(449/258) <= avg(a10) = 1-avg(z10) = 1-r/3 = (3-r)/3 and Z^1 clusters provide an even tighter bound S(449/258) <= avg(a11) = 1-avg(z11) = 1-(5r-4)/8 = (12-5r)/8. The 449/258 example uses 14 Z^0 clusters = ([z10*3])*14 ( 42 z10 pieces, 14 3-parts) 30 Z^1 clusters = (A^0,[z0,z11*2]*4)*30 (240 z11 pieces, 120 3-parts) As before, we refine the labelling of a1 pieces into a10 and a11 pieces, according to z10 or z11 complements. The next level of clustering is constructed by taking closure under a10/z10 muffins. The 282 a1 pieces in the 47 A^1 clusters are divided into 240 a11 pieces and 42 a10 pieces. Thus at least 5 A^1 clusters must contain only a11 pieces, and with maximally even distribution we have 42 A^1 clusters with 1 a10 piece and 5 a11 pieces. Continuing to refine pieces according to their appearance in A^10 or A^11 clusters, we have A^10 = A^1 = {a110*6, ...} (6 a110 pieces and some other pieces) A^11 = Z^0, (A^1)*3 = (a111*15, ...) with 449/258 consisting of 5 A^10 clusters ( 5 A^1 clusters, 30 a110 pieces) 14 A^11 clusters (42 A^1 clusters, 252 a111 pieces) A^11 clusters provide bound S(449/258) <= avg(a111) = (7r-6)/15 The 449/258 example lies on the intersection of this bound and the next level cluster bound. The next cluster level takes a110/z110 muffin closure and induces clusters Z^10 = Z^1 Z^11 = A^10, (Z^1)*6 with 449/258 consisting of 0 Z^10 clusters 5 Z^11 clusters (i.e., only Z^11 clusters) and bound S(449/258) <= avg(a1111) = 1-avg(z1111) = (73-32r)/42. I conclude with some computer generated additional information on these bounds. If anyone sort of understands this post and would like to use my program as is (i.e., no manual and documentation is mostly my posts), drop me a note and I'll send the C++ source (300 lines). B<4>(r) = (1r-0)/4; S'(r) <= B<4>(r) on (3/2,2/1] tight bound on (3/2,8/5] and at r = (4+8k)/(2+5k) (3/2,8/5] contains mediant(3/2,8/5) = 11/7 S'( (3/2,8/5] ) = (3/8,2/5]; S'(11/7) = 11/28 B<>(r) = (1-0r)/2 B<4>(r) = B<>(r) at r = 2/1 = 2.0000000 S'(2/1) = 1/2 = 0.50000000 B<4,1>(r) = (3-1r)/3; S'(r) <= B<4,1>(r) on [12/7,2/1) tight bound on [9/5,2/1) and at r = (12+9k)/(7+5k) [9/5,2/1) contains mediant(9/5,2/1) = 11/6 S'( [9/5,2/1) ) = [2/5,1/3); S'(11/6) = 7/18 B<4>(r) = (1r-0)/4 B<4,1>(r) = B<4>(r) at r = 12/7 = 1.7142857 S'(12/7) = 3/7 = 0.42857143 B<4,1,1>(r) = (2r-1)/6; S'(r) <= B<4,1,1>(r) on (12/7,7/4] tight bound on (12/7,59/34] and at r = (21+59k)/(12+34k) (12/7,59/34] contains mediant(12/7,7/4) = 19/11 S'( (12/7,59/34] ) = (17/42,7/17]; S'(19/11) = 9/22 B<4,1>(r) = (3-1r)/3 B<4,1,1>(r) = B<4,1>(r) at r = 7/4 = 1.7500000 S'(7/4) = 5/12 = 0.41666667 B<4,1,1,1>(r) = (12-5r)/8; S'(r) <= B<4,1,1,1>(r) on [40/23,7/4) tight bound on [148/85,7/4) and at r = (80+148k)/(46+85k) [148/85,7/4) contains mediant(47/27,7/4) = 54/31 S'( [148/85,7/4) ) = [7/17,13/32); S'(54/31) = 51/124 B<4,1,1>(r) = (2r-1)/6 B<4,1,1,1>(r) = B<4,1,1>(r) at r = 40/23 = 1.7391304 S'(40/23) = 19/46 = 0.41304348 B<4,1,1,1,1>(r) = (7r-6)/15; S'(r) <= B<4,1,1,1,1>(r) on (40/23,228/131] tight bound on (40/23,1119/643] and at r = (228+1119k)/(131+643k) (40/23,1119/643] contains mediant(40/23,47/27) = 87/50 S'( (40/23,1119/643] ) = (142/345,265/643]; S'(87/50) = 103/250 B<4,1,1,1>(r) = (12-5r)/8 B<4,1,1,1,1>(r) = B<4,1,1,1>(r) at r = 228/131 = 1.7404580 S'(228/131) = 54/131 = 0.41221374 B<4,1,1,1,1,1>(r) = (73-32r)/42; S'(r) <= B<4,1,1,1,1,1>(r) on [449/258,228/131) tight bound on [9349/5372,228/131) and at r = (1347+9349k)/(774+5372k) [9349/5372,228/131) contains mediant(134/77,47/27) = 181/104 S'( [9349/5372,228/131) ) = [1107/2686,2267/5502); S'(181/104) = 75/182 B<4,1,1,1,1>(r) = (7r-6)/15 B<4,1,1,1,1,1>(r) = B<4,1,1,1,1>(r) at r = 449/258 = 1.7403101 S'(449/258) = 319/774 = 0.41214470 B<4,1,1,1,1,1,1>(r) = (61r-60)/112; S'(r) <= B<4,1,1,1,1,1,1>(r) on (449/258,764/439] tight bound on (449/258,149516/85913] and at r = (10696+149516k)/(6146+85913k) (449/258,149516/85913] contains mediant(449/258,764/439) = 1213/697 S'( (449/258,149516/85913] ) = (11909/28896,35408/85913]; S'(1213/697) = 32173/78064 B<4,1,1,1,1,1>(r) = (73-32r)/42 B<4,1,1,1,1,1,1>(r) = B<4,1,1,1,1,1>(r) at r = 764/439 = 1.7403189 S'(764/439) = 2533/6146 = 0.41213798 B<4,1,1,1,1,1,1,1>(r) = (1101-487r)/615; S'(r) <= B<4,1,1,1,1,1,1,1>(r) on [160212/92059,764/439) tight bound on [1313469/754729,764/439) and at r = (160212+6567345k)/(92059+3773645k) [1313469/754729,764/439) contains mediant(19549/11233,764/439) = 20313/11672 S'( [1313469/754729,764/439) ) = [1555262/3773645,111271/269985); S'(20313/11672) = 986147/2392760 B<4,1,1,1,1,1,1>(r) = (61r-60)/112 B<4,1,1,1,1,1,1,1>(r) = B<4,1,1,1,1,1,1>(r) at r = 160212/92059 = 1.7403187 S'(160212/92059) = 37941/92059 = 0.41213787

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Re: [math-fun] Another(?) stick length illusion
by Dan Asimov 25 Aug '15

25 Aug '15

Huh? At least to the nearest pixel, both lines are the same length. They are corresponding diagonals of congruent regular pentagons. I find this to be an impressive illusion. --Dan << << To me the stick that is in fact longer looks like it is in fact longer. If they're supposed to be the same length I think your image needs a bit of editing. >> On Thu, Aug 19, 2010 at 4:31 PM, Bill Gosper <billgosper(a)gmail.com> wrote: > While hacking the pentagon fill, tutor Julian found > www.tweedledum.com/rwg/6pents.png, which seems to work backwards: > The stick larger than its context appears smaller than the > stick smaller than its context. >> _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] parametrizing the Koch snowflake curve
by Bill Gosper 24 Apr '15

24 Apr '15

Fourier epicycles thru nth: (d16) sum(((product((1 - (( - 1)^j * tan(((%pi * (2 * k + (1/3)))/(2^j)))/(sqrt(3)))),j,1,inf)) * (%e)^(%i * (6 * k + 1) * t)/((2 * k + (1/3))^2)),k, - n,n) See http://www.tweedledum.com/rwg/cog.htm Also http://gosper.org/fst.dvi Pix: http://gosper.org/fst1.png ,http://gosper.org/fst2.png , http://gosper.org/fst3.png ,http://gosper.org/fst4.png . (For parametric, take Re and Im.) Rich's base 4 recursion loops for rational arg. Find the fixed point of the pending stack ops to simulate infinite recursion, and get an exact val on the curve. Equate to this Fourier series for a *non-obvious* identity. --rwg > If you view the starting triangle as running from 0 to 3, > and write the fraction base 4, you can use the > fraction digits to select which part of the (sub)edge > you are on. The orientation of the edge can be calculated > from the 1s and 2s of the high order portion of the fraction, > and that should give a somewhat kludgy sum for the (x,y) > coordinates of the image point. > RWG likely has some nice FFT clock sum expression. > > Rich > > > -----Original Message----- > From: math-fun-bounces(a)mailman.xmission.com [mailto: math-fun-bounces(a)mailman.xmission.com] On > Behalf Of N. J. A. Sloane > Sent: Thursday, February 25, 2010 12:43 PM > To: math-fun(a)mailman.xmission.com > Cc: njas(a)research.att.com > Subject: Re: [math-fun] parametrizing the Koch snowflake curve > > Jim Propp asked: > >> Does anyone (Bill Gosper, maybe?) know an explicit parametrization > of the Koch snowflake curve? > > The Cantor set consists of the points 0 <= x <= 1 which have a base 3 "decimal" expansion that > does not contain a 1. Is there a similar characterization of the Koch snowflake? > > Neil Sloane > >

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[math-fun] why can a line of slope 10^n determine n decimal digits of π?
by Gary Antonick 19 Jun '14

19 Jun '14

Just noticed this yesterday and can't figure out why it works. Draw a circle with radius 1 around 0,0. Draw a line with slope -10 that hits the circle at 1,0. This line will also hit the circle again and make a little arc. The length of the arc will be about 1/31 the perimeter of the circle. Now try again with a line of slope -100. The line will create an arc 1/314 of the circle. Starting to look a bit like pi. Am wondering why this is. In case I made a mistake here's what I'm trying to describe.. if the circle were a clock the first line would cut an arc from 15 minutes (3:00) to almost the 13 minute mark.

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[math-fun] sqrt(pi)
by Bill Gosper 24 Mar '13

24 Mar '13

Pairwise combining the terms of A&S 6.5.31 (upper Gamma CF), then adding last week's lower gamma CF: In[185]:= Equal[Gamma[ z], ((y^z*((1/(cfk[n*(z - n), -z + y + 2*n + 1, List[n, Infinity]] - z + y + 1)) + (1/(z + cfk[n*y, z - y + n, List[n, Infinity]] - y))))/(E^y))] z 1 Gamma[z] == (y (------------------------------------------- + -y + z + cfk[n y, n - y + z, {n, Infinity}] 1 y -----------------------------------------------------------)) / E 1 + y - z + cfk[n (-n + z), 1 + 2 n + y - z, {n, Infinity}] where cfk:=ContinuedFractionK and y>0 can be chosen to optimize convergence. Specializing, In[186]:= % /. z -> 1/2 /. y -> 1 Out[186]= 1 Sqrt[Pi] == (-------------------------------------- + 1 1 -(-) + cfk[n, -(-) + n, {n, Infinity}] 2 2 1 ------------------------------------------) / E 3 1 3 - + cfk[(- - n) n, - + 2 n, {n, Infinity}] 2 2 2 In[192]:= N[List @@ %186 /. cfk[num_, den_, _] :> ContinuedFractionK[num, den, {n, 9}]] Out[192]= {1.77245, 1.77244} This is the first non-asymptotic sum or CF I've seen for sqrt(pi). But the sum behind the lowergamma cf has been there all along, so I must not have been paying attention. --rwg

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Re: [math-fun] Disk polyominoes
by Allan Wechsler 13 Apr '12

13 Apr '12

A funster who might or might not wish to remain anonymous has corrected me in private. A3 does not equal 2; it equals 1. The corrected sequence is: 1, 1, 1, 1, 2, 2, 2, 1, 2, 2, 3, 3 ... As before, I'm not entirely confident of the last couple of values. Thank you for the correction.

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[math-fun] Machine Go
by rcs＠xmission.com 12 Oct '11

12 Oct '11

Science Daily has a report of a recent success for a Go program. http://www.sciencedaily.com/releases/2009/05/090514083931.htm The program Huygens defeated two professional Go players. The program had handicaps of 6 or 7 stones. Question for the Go players among us: Does a handicap of 6 stones mean that the starting board has 6 stones added for the weaker player, or that 6 is added to the final score? Rich

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[math-fun] Slanted coverage
by Bill Gosper 08 Jul '11

08 Jul '11

In those ubiquitous little tables of area and volume formulae in handbooks and textbooks, there's often a picture of a skewed cone next to the right cone to emphasize that they have the same volume formula. Did you ever wonder why there's never the skewed cone area formula? For the special case with the apex directly above a point on the circumference, I get pi h r 3F2[-1/2,1/4,3/4; 1/2, 1; -4 r^2/h^2], which doesn't even converge for h < 2 r. The area element for the general x,y,z parametric surface is just Out[85]= (1,0) (0,1) Norm[Cross[({x, y, z}) , ({x, y, z}) ]] (I don't recall this from 18.03.) E.g., for the torus with radii r and R, In[116]:= % /. {x :> (Cos[#2]*(R + r*Sin[#1]) &), y :> (Sin[#2]*(R + r*Sin[#1]) &), z :> (r*Cos[#1] &)} Out[116]= Norm[Cross[({Cos[#2] (R + r Sin[#1]) & , Sin[#2] (R + r Sin[#1]) & , (1,0) r Cos[#1] & }) , ({Cos[#2] (R + r Sin[#1]) & , (0,1) Sin[#2] (R + r Sin[#1]) & , r Cos[#1] & }) ]] In[117]:= % /. f_[L_List] :> f /@ L /. Abs[w_] -> w Out[117]= Sqrt[Power[-(-r Sin[#1] & ) (Cos[#2] (R + r Sin[#1]) & ) + (0 & ) (r Cos[#1] Sin[#2] & ), 2] + Power[-(0 & ) (r Cos[#1] Cos[#2] & ) + (-r Sin[#1] & ) (-(R + r Sin[#1]) Sin[#2] & ), 2] + Power[(r Cos[#1] Cos[#2] & ) (Cos[#2] (R + r Sin[#1]) & ) - (r Cos[#1] Sin[#2] & ) (-(R + r Sin[#1]) Sin[#2] & ), 2]] In[119]:= % /. (f_ &) :> (f &)[t, u] Out[119]= 2 2 2 2 Sqrt[r Cos[u] Sin[t] (R + r Sin[t]) + 2 2 2 2 r Sin[t] (R + r Sin[t]) Sin[u] + 2 2 2 (r Cos[t] Cos[u] (R + r Sin[t]) + r Cos[t] (R + r Sin[t]) Sin[u] ) ] In[120]:= FullSimplify[%] Out[120]= 2 2 Sqrt[r (R + r Sin[t]) ] In[125]:= Assuming[R > r > 0, Integrate[%120, {t, 0, 2*Pi}, {u, 0, 2*Pi}]] Out[125]= 2 4 Pi r R Did Archimedes know this? --rwg

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[math-fun] Masters of Math, From Old Babylon
by Henry Baker 05 Dec '10

05 Dec '10

FYI -- November 26, 2010 Masters of Math, From Old Babylon By EDWARD ROTHSTEIN If the cost of digging a trench is 9 gin, and the trench has a length of 5 ninda and is one-half ninda deep, and if a workerÂs daily load of earth costs 10 gin to move, and his daily wages are 6 se of silver, then how wide is the canal? Or, a better question: if you were a tutor of Babylonian scribes some 4,000 years ago, holding a clay tablet on which this problem was incised with cuneiform indentations Â the very tablet that can now be seen with 12 others from that Middle Eastern civilization at the Institute for the Study of the Ancient World Â what could you take for granted, and what would you need to explain to your students? In what way did you think about measures of time and space? How did you calculate? Did you believe numbers had an abstract existence, each with its own properties? And how would you have figured out the width of that canal (which, the tablet tells us, is one-and-a-half ninda)? Spend some time at this modest yet thoroughly intriguing exhibition, ÂBefore Pythagoras: The Culture of Old Babylonian Mathematics,Â and you begin to realize that the answers can be far more cryptic than these tablets were before great scholars like Otto E. Neugebauer began to decipher them during the first half of the 20th century. The institute, part of New York University, has gathered together a remarkable selection of Old Babylonian tablets from the collections of three universities Â Columbia, Yale and the University of Pennsylvania Â that cover a wide mathematical range. Made between 1900 and 1700 B.C., they include student exercises, word problems and calculation tables, as well as more abstract demonstrations. Under the curatorship of Alexander Jones, a professor at N.Y.U., and Christine Proust, a historian of mathematics, the tablets are used to give a quick survey of Babylonian mathematical enterprise, while also paying tribute to Neugebauer, the Austrian-born scholar who spent the last half of his career teaching at Brown University and almost single-handedly created a new discipline of study through his analysis of these neglected sources. Only about 950 mathematically oriented tablets survived two millenniums of Babylonian history, and since their discovery, debate has raged over what they show us about that lost world. Every major history of Western mathematics written during the last 70 years has at least started to take Babylonians into account. Generally, their systems have been seen as precursors to the theoretical flowering of Greek mathematics, out of which our own mathematical approaches have grown. But Neugebauer, and then his many students and rivals, also showed how sophisticated Babylonian mathematics was and how many similarities existed to later Western systems Â if, that is, you counted using 60 fingers (as we often seem to, thanks to the Babylonians, when dealing with seconds and minutes and, in part, even when measuring angles). Examining the surviving tablets, including one multiplication table on display here, scholars decoded the birdÂs feet of Babylonian numerals, showing that the Babylonians, like us, used the same symbol to represent different numerical values. (The same digit for us has a different value if it is in the 1Âs column, the 10Âs column, or the 100Âs column; the Babylonians could use the same sign, depending on context, to represent a 1 or a 60 or a 3,600.) The most famous tablets here Â one showing a square with two diagonals, and another, known as Plimpton 322, containing a table of numerical symbols Â suggest that the Babylonians knew at least some of the consequences of the theorem that now bears the name of the Greek mathematician Pythagoras, who lived some 1,500 years after these works were chiseled. But did the Babylonians conceive of it as a ÂtheoremÂ Â a timeless truth subject to proof based on accepted principles? Or was it thought of as a property of areas of land that were mapped out by surveyors? Or, as one scholar recently wrote, was Plimpton 322 Âa teacherÂs catalog of parametersÂ for calculation? Or something else completely? Aside from the fact that the analysis of these tablets is relatively recent, one of the problems is that much of their context is hypothetical, because of the almost haphazard way in which early modern explorers pulled these artifacts out of the layered rubble of ancient mounds of detritus. In a fascinating 2008 book, ÂMathematics in Ancient Iraq: A Social HistoryÂ (Princeton), Eleanor Robson even suggests that many tablets like these of the second millennium B.C. Âwere essentially ephemera, created to aid and demonstrate recall, destined almost immediately for the recycling bin.Â But as Ms. Robson also points out, these tabletsÂ word problems about digging and construction, their use in teaching record keeping and calculation, and their implicit affirmation of the importance of scribes and teachers, also reveal a highly organized, bureaucratic society, an Âordered urban state, with god, king and scribe at its center.Â At the exhibition itself, it would have helped if some of the translations and interpretations of the tablets provided in an accompanying brochure had been made available directly on the labels themselves, so you could both look at the artifact and see how to interpret it. It might have helped, too, if the show suggested the kinds of debates that have arisen about the status of Babylonian mathematics, which, in many ways, parallel larger debates about Western and non-Western cultures. In this case those debates become all the more charged because of the geographical region involved. While once, for example, Babylonian mathematics was clearly seen as a precursor to the Greeks and to Western systems, there is an impulse now to see it as a historical victim of Western cultural perspectives. That is partly how Ms. Robson portrays it, even (unconvincingly) evoking Edward SaidÂs book ÂOrientalismÂ; a hint of that sentiment may even be one of the subtle draws of this exhibition. But for an outsider surveying these objects and the history, the response is at once awe at the ancient minds that created such powerful systems for understanding and ordering experience, and still more amazement at the mathematical world that later developed and went so much further: explicitly turning individual examples into theorems, earthly practices into abstract principles and outlining new methods for understanding that are still being applied. This is the trench solution as it appears on the Babylonian tablet. The translation, by Alexander Jones, a curator of the ÂBefore PythagorasÂ exhibition, is in the showÂs brochure: ÂSolution: Multiply the length and the depth, and you get 30. Take the reciprocal of the workload, multiply by 30 and you will get 3. Multiply the wages by 3, and you will get 6. Take the reciprocal of 6, and multiply it by 9, the total cost in silver, and you will get its width. One and a half ninda is the width. Such is the procedure.Â (A reciprocal in Babylonian arithmetic is in relation to 60, so the reciprocal of 10 is 6.) ÂBefore Pythagoras: The Culture of Old Babylonian MathematicsÂ is on view through Dec. 17 at the Institute for the Study of the Ancient World, New York University, 15 East 84th Street, Manhattan; nyu.edu/isaw.

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[math-fun] Integers shaped like right triangles
by Eric Angelini 01 Dec '10

01 Dec '10

Hello Math-Fun, Write an integer somewhere on a grid, one digit per square: +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | 1 | 3 | 4 | 0 | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ The digits will now move vertically: - odd digit j will climb j squares - even digit k will sink k squares - zero stays were it is In our example, the digits of 1340 move like this: +---+---+---+---+ | | 3 | | | +---+---+---+---+ | | | | | +---+---+---+---+ | 1 | | | | +---+---+---+---+ | . | . | . | 0 | +---+---+---+---+ | | | | | +---+---+---+---+ | | | | | +---+---+---+---+ | | | | | +---+---+---+---+ | | | 4 | | +---+---+---+---+ Encircle now those new digits with an elastic band. The resulting shape is not a right triangle - which is a pity because we are precisely looking for integers moving into right triangles! The additional constraint being that ALL the new digits must be ON the elastic band. [you can see this more clearly there: http://www.cetteadressecomportecinquantesignes.com/RightTriangleNumbers.htm ] Here are a few good examples: Integer 130 (the right angle is on 1): +---+---+---+---+ | | 3 | | | +---+---+---+---+ | | | | | +---+---+---+---+ | 1 | | | | +---+---+---+---+ | . | . | 0 | | +---+---+---+---+ Integer 313333 (right angle on 1): +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | 3 | | 3 | 3 | 3 | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | 1 | | | | | +---+---+---+---+---+---+ | . | . | . | . | . | | +---+---+---+---+---+---+ Integer 2201022 (right angle on 1): +---+---+---+---+---+---+---+ | | | | | | | | +---+---+---+---+---+---+---+ | | | | 1 | | | | +---+---+---+---+---+---+---+ | . | . | 0 | . | 0 | . | . | +---+---+---+---+---+---+---+ | | | | | | | | +---+---+---+---+---+---+---+ | 2 | 2 | | | | 2 | 2 | +---+---+---+---+---+---+---+ | | | | | | | | +---+---+---+---+---+---+---+ Here is a bad example: Integer 21000 (yes we have a triangle, yes there is a right angle --on 1--, but the band doesn't pass through ALL digits): +---+---+---+---+---+---+---+ | | | | | | | | +---+---+---+---+---+---+---+ | | | 1 | | | | | +---+---+---+---+---+---+---+ | | . | . | 0 | 0 | 0 | | +---+---+---+---+---+---+---+ | | | | | | | | +---+---+---+---+---+---+---+ | | 2 | | | | | | +---+---+---+---+---+---+---+ | | | | | | | | +---+---+---+---+---+---+---+ Questions: a) Find the 40 first integers producing a right triangle b) Find the biggest integer producing a right triangle c) Find the smallest integer producing a rectangle. [Answers here in December 2010] Best, É.

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