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Frequently Asked Questions (FAQS);faqs.452 ==> probability/roulette.p <== You are in a game of Russian roulette, but this time the gun (a 6 shooter revolver) has three bullets _in_a_row_ in three of the chambers. The barrel is spun only once. Each player then points the gun at his (her) head and pulls the trigger. If he (she) is still ==> probability/unfair.p <== Generate even odds from an unfair coin. For example, if you thought a coin was biased toward heads, how could you get the equivalent of a fair coin with several tosses of the unfair coin? ==> series/series.01.p <== M, N, B, D, P ? ==> series/series.02.p <== H, H, L, B, B, C, N, O, F ? ==> series/series.03.p <== W, A, J, M, M, A, J? ==> series/series.03a.p <== G, J, T, J, J, J, A, M, W, J, J, Z, M, F, J, ? ==> series/series.03b.p <== A, J, B, C, G, T, C, V, J, T, D, F, K, B, H, ? ==> series/series.03c.p <== M, A, M, D, E, L, R, H, ? ==> series/series.04.p <== A, E, H, I, K, L, ? ==> series/series.05.p <== A B C D E F G H? ==> series/series.06.p <== Z, O, T, T, F, F, S, S, E, N? ==> series/series.06a.p <== F, S, T, F, F, S, ? ==> series/series.07.p <== 1, 1 1, 2 1, 1 2 1 1, ... What is the pattern and asymptotics of this series? ==> series/series.08a.p <== G, L, M, B, C, L, M, C, F, S, ? ==> series/series.08b.p <== A, V, R, R, C, C, L, L, L, E, ? ==> series/series.09a.p <== S, M, S, S, S, C, P, P, P, ? ==> series/series.09b.p <== M, S, C, P, P, P, S, S, S, ? ==> series/series.10.p <== D, P, N, G, C, M, M, S, ? ==> series/series.11.p <== R O Y G B ? ==> series/series.12.p <== A, T, G, C, L, ? ==> series/series.13.p <== M, V, E, M, J, S, ? ==> series/series.14.p <== A, B, D, O, P, ? ==> series/series.14a.p <== A, B, D, E, G, O, P, ? ==> series/series.15.p <== A, E, F, H, I, ? ==> series/series.16.p <== A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, X, Y? ==> series/series.17.p <== T, P, O, F, O, F, N, T, S, F, T, F, E, N, S, N? ==> series/series.18.p <== 10, 11, 12, 13, 14, 15, 16, 17, 20, 22, 24, ___ , 100, 121, 10000 ==> series/series.19.p <== 1 01 01011 0101101011011 0101101011011010110101101101011011 etc. Each string is formed from the previous string by substituting '01' for '1' and '011' for '0' simultaneously at each occurance. ==> series/series.20.p <== 1 2 5 16 64 312 1812 12288 ==> series/series.21.p <== 5, 6, 5, 6, 5, 5, 7, 5, ? ==> series/series.22.p <== 3 1 1 0 3 7 5 5 2 ? ==> series/series.23.p <== 22 22 30 13 13 16 16 28 28 11 ? ==> series/series.24.p <== What is the next letter in the sequence: W, I, T, N, L, I, T? ==> series/series.25.p <== 1 3 4 9 10 12 13 27 28 30 31 36 37 39 40 ? ==> series/series.26.p <== 1 3 2 6 7 5 4 12 13 15 14 10 11 9 8 24 25 27 26 ? ==> series/series.27.p <== 0 1 1 2 1 2 1 3 2 2 1 3 1 2 2 4 1 3 1 3 2 2 1 4 2 ? ==> series/series.28.p <== 0 2 3 4 5 5 7 6 6 7 11 7 13 9 8 8 17 8 19 9 10 13 23 9 10 ? ==> series/series.29.p <== 1 1 2 1 2 2 3 1 2 2 3 2 3 3 4 1 2 2 3 2 3 3 4 2 3 3 4 3 4 ? ==> series/series.30.p <== I I T Y W I M W Y B M A D ==> series/series.31.p <== 6 2 5 5 4 5 6 3 7 ==> series/series.32.p <== 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1 ==> series/series.33.p <== 2 12 360 75600 ==> series/series.34.p <== 3 5 4 4 3 5 5 4 3 ==> series/series.35.p <== 1 2 3 2 1 2 3 4 2 1 2 3 4 2 2 3 ==> trivia/area.codes.p <== When looking at a map of the distribution of telephone area codes for North America, it appears that they are randomly distributed. I am doubtful that this is the case, however. Does anyone know how the area codes were/are chosen? ==> trivia/eskimo.snow.p <== How many words do the Eskimo have for snow? ==> trivia/federal.reserve.p <== What is the pattern to this list: Boston, MA New York, NY Philadelphia, PA ==> trivia/jokes.self-referential.p <== What are some self-referential jokes? Xref: bloom-picayune.mit.edu rec.puzzles:18137 news.answers:3069 Newsgroups: rec.puzzles,news.answers Path: bloom-picayune.mit.edu!enterpoop.mit.edu!snorkelwacker.mit.edu!usc!wupost!uunet!questrel!chris From: uunet!questrel!chris (Chris Cole) Subject: rec.puzzles FAQ, part 2 of 15 Message-ID: <puzzles-faq-2_717034101@questrel.com> Followup-To: rec.puzzles Summary: This posting contains a list of Frequently Asked Questions (and their answers). It should be read by anyone who wishes to post to the rec.puzzles newsgroup. Sender: chris@questrel.com (Chris Cole) Reply-To: uunet!questrel!faql-comment Organization: Questrel, Inc. References: <puzzles-faq-1_717034101@questrel.com> Date: Mon, 21 Sep 1992 00:08:31 GMT Approved: news-answers-request@MIT.Edu Expires: Sat, 3 Apr 1993 00:08:21 GMT Lines: 1579 Archive-name: puzzles-faq/part02 Last-modified: 1992/09/20 Version: 3 ==> analysis/bugs.p <== Four bugs are placed at the corners of a square. Each bug walks directly toward the next bug in the clockwise direction. The bugs walk with constant speed always directly toward their clockwise neighbor. Assuming the bugs make at least one full circuit around the center of the square before meeting, how much closer to the center will a bug be at the end of its first full circuit? ==> analysis/bugs.s <== Amorous Bugs ANSWER: 1 - e^(-2*pi) Let O(t) be the angle at time t of bug 1 relative to its starting point and r(O(t)) be its distance from the center of the square. Bug 1's vector trajectory is (using a Cartesian coordinate system with the origin at the center of the square): (1) X1 = [r(O) * cos(O), r(O) * sin(O)] By symmetry, bug 2's trajectory is the same only rotated by pi/2, viz.: (2) X2 = [-r(O) * sin(O), r(O) * cos(O)] Since bug 1 walks directly toward bug 2, the velocity of bug 1 must be proportional to the vector from bug 1 to bug 2: (3) d(X1)/d(t) = k * (X2 - X1) Equating each component of the vector equation (3) yields: (4) (d(r)/d(O) * cos(O) - r * sin(O)) * d(O)/d(t) = k * (-r * cos(O) - r * sin(O)) (5) (d(r)/d(O) * sin(O) + r * cos(O)) * d(O)/d(t) = k * (-r * sin(O) + r * cos(O)) These equations are solved by: (6) k = d(O)/d(t) and: (7) d(r)/d(O) = -r(O) (7) is solved by: (8) r(O) = e^-O Constant speed gives: (9) v^2 = constant = ((d(r)/d(O))^2+r^2)*(d(O)/d(t))^2 Substituting (8) into (9) yields (let V = v/sqrt(2)): (10) d(O)/d(t) = V * e^O Which is solved (using the boundary condition O(0) = 0) by: (11) O(t) = -ln(1 - V * t) Substituting (11) into (8) yields: (12) r(t) = r(0) - V * t The bug has made a full circle when O(T) = 2*pi; using (11): (13) T = 1/V * (1 - e^(-2*pi)) Substituting T into (12) yields the answer: (14) r(T) - r(0) = 1 - e^(-2*pi) ==> analysis/c.infinity.p <== What function is zero at zero, strictly positive elsewhere, infinitely differentiable at zero and has all zero derivitives at zero? ==> analysis/c.infinity.s <== exp(-1/x^2) This tells us why Taylor Series are a more limited device than they might be. We form a Taylor series by looking at the derivatives of a function at a given point; but this example shows us that the derivatives at a point may tell us almost nothing about its behavior away from that point. ==> analysis/cache.p <== Cache and Ferry (How far can a truck go in a desert?) A pick-up truck is in the desert beside N 50-gallon gas drums, all full. The truck's gas tank holds 10 gallons and is empty. The truck can carry one drum, whether full or empty, in its bed. It gets 10 miles to the gallon. How far away from the starting point can you drive the truck? ==> analysis/cache.s <== If the truck can siphon gas out of its tank and leave it in the cache, the answer is: { 1/1 + 1/3 + ... + 1/(2 * N - 1) } x 500 miles. Otherwise, the "Cache and Ferry" problem is the same as the "Desert Fox" problem described, but not solved, by Dewdney, July '87 "Scientific American". Dewdney's Oct. '87 Sci. Am. article gives for N=2, the optimal distance of 733.33 miles. In the Nov. issue, Dewdney lists the optimal distance of 860 miles for N=3, and gives a better, but not optimal, general distance formula. Westbrook, in Vol 74, #467, pp 49-50, March '90 "Mathematical Gazette", gives an even better formula, for which he incorrectly claims optimality: For N = 2,3,4,5,6: Dist = (600/1 + 600/3 + ... + 600/(2N-3)) + (600-100N)/(2N-1) For N > 6: Dist = (600/1 + 600/3 + ... + 600/9) + (500/11 + ... + 500/(2N-3)) The following shows that Westbrook's formula is not optimal for N=8: Ferry 7 drums forward 33.3333 miles (356.6667 gallons remain) Ferry 6 drums forward 51.5151 miles (300.0000 gallons remain) Ferry 5 drums forward 66.6667 miles (240.0000 gallons remain) Ferry 4 drums forward 85.7143 miles (180.0000 gallons remain) Ferry 3 drums forward 120.0000 miles (120.0000 gallons remain) Ferry 2 drums forward 200.0000 miles ( 60.0000 gallons remain) Ferry 1 drums forward 600.0000 miles --------------- Total distance = 1157.2294 miles (Westbrook's formula = 1156.2970 miles) ["Ferrying n drums forward x miles" involves (2*n-1) trips, each of distance x.] Other attainable values I've found: N Distance --- --------- (Ferry distances for each N are omitted for brevity.) 5 1016.8254 7 1117.8355 11 1249.2749 13 1296.8939 17 1372.8577 19 1404.1136 (The N <= 19 distances could be optimal.) 31 1541.1550 (I doubt that this N = 31 distance is optimal.) 139 1955.5509 (I'm sure that this N = 139 distance is not optimal.) So...where's MY formula? I haven't found one, and believe me, I've looked. I would be most grateful if someone would end my misery by mailing me a formula, a literature reference, or even an efficient algorithm that computes the optimal distance. If you do come up with the solution, you might want to first check it against the attainable distances listed above, before sending it out. (Not because you might be wrong, but just as a mere formality to check your work.) [Warning: the Mathematician General has determined that this problem is as addicting as Twinkies.] Myron P. Souris | "If you have anything to tell me of importance, McDonnell Douglas | for God's sake begin at the end." souris@mdcbbs.com | Sara Jeanette Duncan @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ The following output comes from some hack programs that I've used to empirically verify some proofs I've been working on. Initial barrels: 12 (600 gallons) Attainable distance= 1274.175211 Barrels Distance Gas Moved covered left >From depot 1: 10 63.1579 480.0000 >From depot 2: 8 50.0000 405.0000 >From depot 3: 7 37.5000 356.2500 >From depot 4: 6 51.1364 300.0000 >From depot 5: 5 66.6667 240.0000 >From depot 6: 4 85.7143 180.0000 >From depot 7: 3 120.0000 120.0000 >From depot 8: 2 200.0000 60.0000 >From depot 9: 1 600.0000 0.0000 Initial barrels: 40 (2000 gallons) Attainable distance= 1611.591484 Barrels Distance Gas Moved covered left >From depot 1: 40 2.5316 1980.0000 >From depot 2: 33 50.0000 1655.0000 >From depot 3: 28 50.0000 1380.0000 >From depot 4: 23 53.3333 1140.0000 >From depot 5: 19 50.0000 955.0000 >From depot 6: 16 56.4516 780.0000 >From depot 7: 13 50.0000 655.0000 >From depot 8: 11 54.7619 540.0000 >From depot 9: 9 50.0000 455.0000 >From depot 10: 8 32.1429 406.7857 >From depot 11: 7 38.9881 356.1012 >From depot 12: 6 51.0011 300.0000 >From depot 13: 5 66.6667 240.0000 >From depot 14: 4 85.7143 180.0000 >From depot 15: 3 120.0000 120.0000 >From depot 16: 2 200.0000 60.0000 >From depot 17: 1 600.0000 0.0000 ==> analysis/cats.and.rats.p <== If 6 cats can kill 6 rats in 6 minutes, how many cats does it take to kill one rat in one minute? ==> analysis/cats.and.rats.s <== The following piece by Lewis Carroll first appeared in ``The Monthly Packet'' of February 1880 and is reprinted in _The_Magic_of_Lewis_Carroll_, edited by John Fisher, Bramhall House, 1973. /Larry Denenberg larry@bbn.com larry@harvard.edu Cats and Rats If 6 cats kill 6 rats in 6 minutes, how many will be needed to kill 100 rats in 50 minutes? This is a good example of a phenomenon that often occurs in working problems in double proportion; the answer looks all right at first, but, when we come to test it, we find that, owing to peculiar circumstances in the case, the solution is either impossible or else indefinite, and needing further data. The 'peculiar circumstance' here is that fractional cats or rats are excluded from consideration, and in consequence of this the solution is, as we shall see, indefinite. The solution, by the ordinary rules of Double Proportion, is as follows: 6 rats : 100 rats \ > :: 6 cats : ans. 50 min. : 6 min. / . . . ans. = (100)(6)(6)/(50)(6) = 12 But when we come to trace the history of this sanguinary scene through all its horrid details, we find that at the end of 48 minutes 96 rats are dead, and that there remain 4 live rats and 2 minutes to kill them in: the question is, can this be done? Now there are at least *four* different ways in which the original feat, of 6 cats killing 6 rats in 6 minutes, may be achieved. For the sake of clearness let us tabulate them: A. All 6 cats are needed to kill a rat; and this they do in one minute, the other rats standing meekly by, waiting for their turn. B. 3 cats are needed to kill a rat, and they do it in 2 minutes. C. 2 cats are needed, and do it in 3 minutes. D. Each cat kills a rat all by itself, and take 6 minutes to do it. In cases A and B it is clear that the 12 cats (who are assumed to come quite fresh from their 48 minutes of slaughter) can finish the affair in the required time; but, in case C, it can only be done by supposing that 2 cats could kill two-thirds of a rat in 2 minutes; and in case D, by supposing that a cat could kill one-third of a rat in two minutes. Neither supposition is warranted by the data; nor could the fractional rats (even if endowed with equal vitality) be fairly assigned to the different cats. For my part, if I were a cat in case D, and did not find my claws in good working order, I should certainly prefer to have my one-third-rat cut off from the tail end. In cases C and D, then, it is clear that we must provide extra cat-power. In case C *less* than 2 extra cats would be of no use. If 2 were supplied, and if they began killing their 4 rats at the beginning of the time, they would finish them in 12 minutes, and have 36 minutes to spare, during which they might weep, like Alexander, because there were not 12 more rats to kill. In case D, one extra cat would suffice; it would kill its 4 rats in 24 minutes, and have 24 minutes to spare, during which it could have killed another 4. But in neither case could any use be made of the last 2 minutes, except to half-kill rats---a barbarity we need not take into consideration. To sum up our results. If the 6 cats kill the 6 rats by method A or B, the answer is 12; if by method C, 14; if by method D, 13. This, then, is an instance of a solution made `indefinite' by the circumstances of the case. If an instance of the `impossible' be desired, take the following: `If a cat can kill a rat in a minute, how many would be needed to kill it in the thousandth part of a second?' The *mathematical* answer, of course, is `60,000,' and no doubt less than this would *not* suffice; but would 60,000 suffice? I doubt it very much. I fancy that at least 50,000 of the cats would never even see the rat, or have any idea of what was going on. Or take this: `If a cat can kill a rat in a minute, how long would it be killing 60,000 rats?' Ah, how long, indeed! My private opinion is that the rats would kill the cat. ==> analysis/e.and.pi.p <== Which is greater, e^(pi) or (pi)^e ? ==> analysis/e.and.pi.s <== Put x = pi/e - 1 in the inequality e^x > 1+x (x>0). ==> analysis/functional/distributed.p <== Find all f: R -> R, f not identically zero, such that (*) f( (x+y)/(x-y) ) = ( f(x)+f(y) )/( f(x)-f(y) ). ==> analysis/functional/distributed.s <== 1) Assuming f finite everywhere, (*) ==> x<>y ==> f(x)<>f(y) 2) Exchanging x and y in (*) we see that f(-x) = -f(x). 3) a <> 0 ==> f((a-a)/(a+a)) = (f(a)-f(a))/(f(a)+f(a)) ==> f(0) = 0. 4) a <> 0 ==> f((a+0)/(a-0)) = f(a)/f(a) ==> f(1) = 1. 5) x<>y, y<>0 ==> f(x/y) = f( ((x+y)/(x-y) + (x-y)/(x-y)) / ((x+y)/(x-y) - (x-y)/(x-y)) = f(x)/f(y) ==> f(xy) = f(x)f(y) by replacing x with xy and by noting that f(x*1) = f(x)*1 and f(x*0) = f(x)*f(0). 6) f(x*x) = f(x)*f(x) ==> f(x) > 0 if x>0. 7) Let a=1+\/2, b=1-\/2; a,b satisfy (x+1)/(x-1) = x ==> f(x) = (f(x)+1)/(f(x)-1) ==> f(a)=a, f(b)=b. f(1/\/2) = f((a+b)/(a-b)) = (a+b)/(a-b) = 1/\/2 ==> f(2) = 2. 8) By induction and the relation f((n+1)/(n-1)) = (f(n)+1)/(f(n)-1) we get that f(n)=n for all integer n. #5 now implies that f fixes the rationals. 9) If x>y>0 (*) ==> f(x) - f(y) = f(x+y)/f((x+y)/(x-y)) > 0 by #6. Thus f is order-preserving. Since f fixes the rationals *and* f is order-preserving, f must be the identity function. This was E2176 in _The American Mathematical Monthly_ (the proposer was R. S. Luthar). ==> analysis/functional/linear.p <== Suppose f is non-decreasing with f(x+y) = f(x) + f(y) + C for all real x, y. Prove: there is a constant A such that f(x) = Ax - C for all x. (Note: continuity of f is not assumed in advance.) ==> analysis/functional/linear.s <== By induction f(mx) = m(f(x)+C)-C. Let x=1/n, m=n and find that f(1/n) = (1/n)(f(1)+C)-C. Now let x=1/n and find that f(m/n) = (m/n)(f(1)+C)-C. f(-x+x) = f(-x) + f(x) + C ==> f(-x) = -2C - f(x) (since f(0) = -C) ==> f(-m/n) = -(m/n)(f(1)+C)-C. Since f is monotonic ==> f(x) = x*(f(1)+C)-C for all real x (Squeeze Theorem). ==> analysis/integral.p <== If f is integrable on (0,inf), and differentiable at 0, and a > 0, show: inf ( f(x) - f(ax) ) Int ---------------- dx = f(0) ln(a) 0 x ==> analysis/integral.s <== First, note that if f(0) is 0, then by substituting u=ax in the integral of f(x)/x, our integral is the difference of two equal integrals and so is 0 (the integrals are finite because f is 0 at 0 and differentiable there. Note I make no requirement of continuity). Second, note that if f is the characteristic function of the interval [0, 1]--- i.e. 1, 0<=x<=1 f (x) = 0 otherwise then a little arithmetic reduces our integral to that of 1/x from 1/a to 1 (assuming a>1; if a <= 1 the reasoning is similar), which is ln(a) = f(0)ln(a) as required. Call this function g. Finally, note that the operator which takes the function f to the value of our integral is linear, and that every function meeting the hypotheses (incidentally, I should have said `differentiable from the right', or else replaced the characteristic function of [0,1] above by that of (-infinity, 1]; but it really doesn't matter) is a linear combination of one which is 0 at 0 and g, to wit f(x) = f(0)g(x) + (f(x) - g(x)f(0)). ==> analysis/period.p <== What is the least possible integral period of the sum of functions of periods 3 and 6? ==> analysis/period.s <== Period 2. Clearly, the sum of periodic functions of periods 2 and three is 6. So take the function which is the sum of that function of period six and the negative of the function of period three and you have a function of period 2. ==> analysis/rubberband.p <== A bug walks down a rubberband which is attached to a wall at one end and a car moving away from the wall at the other end. The car is moving at 1 m/sec while the bug is only moving at 1 cm/sec. Assuming the rubberband is uniformly and infinitely elastic, will the bug ever reach the car? ==> analysis/rubberband.s <== Let w = speed of bug and N = ratio of car speed/bug speed = 100. Paint N+1 equally spaced stripes on the rubberband. When the bug is standing on one stripe, the next stripe is moving away from him at a speed slightly < w (relative to him). Since he is walking at w, clearly the bug can reach the next stripe. But once he reaches that stripe, the next one is only receeding at < w. So he walks on down to the car, one stripe at a time. The bug starts gaining on the car when he is at the next to last stripe. ==> analysis/series.p <== Show that in the series: x, 2x, 3x, .... (n-1)x (x can be any real number) there is at least one number which is within 1/n of an integer. ==> analysis/series.s <== Throw 0 into the sequence; there are now n numbers, so some pair must have fractional parts within 1/n of each other; their difference is then within 1/n of an integer. ==> analysis/snow.p <== Snow starts falling before noon on a cold December day. At noon a snowplow starts plowing a street. It travels 1 mile in the first hour, and 1/2 mile in the second hour. What time did the snow start falling?? You may assume that the plow's rate of travel is inversely proportioned to the height of the snow, and that the snow falls at a uniform rate. ==> analysis/snow.s <== 11:22:55.077 am. Method: Let b = the depth of the snow at noon, a = the rate of increase in the depth. Then the depth at time t (where noon is t=0) is at+b, the snowfall started at t_0=-b/a, and the snowplow's rate of progress is ds/dt = k/(at+b). If the snowplow starts at s=0 then s(t) = (k/a) log(1+at/b). Note that s(2 hours) = 1.5 s(1 hour), or log(1+2A/b) = 1.5 log(1+A/b), where A = (1 hour)*a. Letting x = A/b we have (1+2x)^2 = (1+x)^3. Solve for x and t_0 = -(1 hour)/x. The exact answer is 11:(90-30 Sqrt[5]). _American Mathematics Monthly_, April 1937, page 245 E 275. Proposed by J. A. Benner, Lafayette College, Easton. Pa. The solution appears, appropriately, in the December 1937 issue, pp. 666-667. Also solved by William Douglas, C. E. Springer, E. P. Starke, W. J. Taylor, and the proposer. See R.P. Agnew, "Differential Equations," 2nd edition, p. 39 ff. ==> analysis/tower.p <== A number is raised to its own power. The same number is then raised to the power of this result. The same number is then raised to the power of this second result. This process is continued forever. What is the maximum number which will yield a finite result from this process? ==> analysis/tower.s <== Tower of Exponentials ANSWER: e^(1/e) Let N be the number in question and R the result of the process. Then R can be defined recursively by the equation: (1) R = N^R Taking the logarithm of both sides of (1): (2) ln(R) = R * ln(N) Dividing (2) by R and rearranging: (3) ln(N) = ln(R) / R Exponentiating (3): (4) N = R^(1/R) We wish to find the maximum value of N with respect to R. Find the derivative of N with respect to R and set it equal to zero: (5) d(N)/d(R) = (1 - ln(R)) / R^2 = 0 For finite values of R, (5) is satisfied by R = e. This is a maximum of N if the second derivative of N at R = e is less than zero. (6) d2(N)/d2(R) | R=e = (2 * ln(R) - 3) / R^3 | R=e = -1 / e^3 < 0 The solution therefore is (4) at R = e: (7) Nmax = e^(1/e) ==> arithmetic/7-11.p <== A customer at a 7-11 store selected four items to buy, and was told that the cost was $7.11. He was curious that the cost was the same as the store name, so he inquired as to how the figure was derived. The clerk said that he had simply multiplied the prices of the four individual items. The customer protested that the four prices should have been ADDED, not MULTIPLIED. The clerk said that that was OK with him, but, the result was still the same: exactly $7.11. What were the prices of the four items? ==> arithmetic/7-11.s <== The prices are: $1.20, $1.25, $1.50, and $3.16 $7.11 is not the only number which works. Here are the first 160 such numbers, preceded by a count of distinct solutions for that price. Note that $7.11 has a single, unique solution. 1 - $6.44 1 - $7.83 2 - $9.20 3 - $10.89 1 - $6.51 1 - $7.86 1 - $9.23 1 - $10.95 1 - $6.60 3 - $7.92 1 - $9.24 2 - $11.00 1 - $6.63 1 - $8.00 1 - $9.27 1 - $11.07 1 - $6.65 1 - $8.01 2 - $9.35 1 - $11.13 1 - $6.72 1 - $8.03 3 - $9.36 1 - $11.16 2 - $6.75 5 - $8.10 1 - $9.38 1 - $11.22 1 - $6.78 1 - $8.12 5 - $9.45 2 - $11.25 1 - $6.80 1 - $8.16 2 - $9.48 2 - $11.27 2 - $6.84 2 - $8.19 1 - $9.54 1 - $11.30 1 - $6.86 1 - $8.22 1 - $9.57 1 - $11.36 1 - $6.89 1 - $8.25 1 - $9.59 1 - $11.40 2 - $6.93 3 - $8.28 2 - $9.60 2 - $11.43 1 - $7.02 3 - $8.33 1 - $9.62 2 - $11.52 1 - $7.05 1 - $8.36 2 - $9.63 2 - $11.55 2 - $7.07 1 - $8.37 1 - $9.66 2 - $11.61 1 - $7.08 2 - $8.40 1 - $9.68 1 - $11.69 1 - $7.11 1 - $8.45 2 - $9.69 1 - $11.70 1 - $7.13 2 - $8.46 1 - $9.78 1 - $11.88 2 - $7.14 1 - $8.52 2 - $9.80 1 - $11.90 3 - $7.20 5 - $8.55 1 - $9.81 1 - $11.99 1 - $7.25 1 - $8.60 1 - $9.87 1 - $12.06 1 - $7.26 4 - $8.64 4 - $9.90 1 - $12.15 2 - $7.28 1 - $8.67 1 - $9.92 1 - $12.18 1 - $7.29 1 - $8.69 2 - $9.99 1 - $12.24 3 - $7.35 1 - $8.73 1 - $10.01 1 - $12.30 1 - $7.37 2 - $8.75 1 - $10.05 1 - $12.32 1 - $7.47 1 - $8.76 2 - $10.08 1 - $12.35 1 - $7.50 1 - $8.78 1 - $10.17 2 - $12.42 1 - $7.52 5 - $8.82 1 - $10.20 1 - $12.51 4 - $7.56 1 - $8.85 2 - $10.26 1 - $12.65 1 - $7.62 1 - $8.88 3 - $10.29 2 - $12.69 4 - $7.65 2 - $8.91 3 - $10.35 1 - $12.75 1 - $7.67 1 - $8.94 2 - $10.44 1 - $12.92 2 - $7.70 1 - $8.96 1 - $10.53 1 - $12.96 3 - $7.74 3 - $9.00 1 - $10.56 1 - $13.23 1 - $7.77 1 - $9.02 1 - $10.64 1 - $13.41 1 - $7.79 2 - $9.03 2 - $10.71 1 - $13.56 2 - $7.80 1 - $9.12 3 - $10.80 1 - $14.49 1 - $7.82 2 - $9.18 1 - $10.85 1 - $15.18 There are plenty of solutions for five summands. Here are a few: $8.28 -- at least two solutions $8.47 -- at least two solutions $8.82 -- at least two solutions -- Mark Johnson mark@microunity.com (408) 734-8100 There may be many approximate solutions, for example: $1.01, $1.15, $2.41, and $2.54. These sum to $7.11 but the product is 7.1100061. ==> arithmetic/clock/day.of.week.p <== It's restful sitting in Tom's cosy den, talking quietly and sipping a glass of his Madeira. I was there one Sunday and we had the usual business of his clock. When the radio gave the time at the hour, the Ormolu antique was exactly 3 minutes slow. "It loses 7 minutes every hour", my old friend told me, as he had done so many times before. "No more and no less, but I've gotten used to it that way." When I spent a second evening with him later that same month, I remarked on the fact that the clock was dead right by radio time at the hour. It was rather late in the evening, but Tom assured me that his treasure had not been adjusted nor fixed since my last visit.