Charles has been frustrated at every turn, and a waiter suggested the following, succinct, problem for Maggie (and for himself!):

Here's one by a former student, Roy Kilgard, who's listed below among the entries:

From: Roy KilgardCome now, Barnette, this one's easy!!! To time nine minutes with a four minute and seven minute timer: Start both timers simultaneously. When the four minute runs out, start it again. When the seven minute runs out, begin timing nine minute interval. Let the remaining minute fall through the four minute and run it twice more. 7-4=3 4-3=1 1+4+4=9 Simple enough. -Roy

THANKS TO ALL THE ENTRANTS! KEEP IT UP!

From: Danny SharpeHi, Ron. You probably don't remember me but I took a couple of courses from you back around 1980 +/-. Found your departmental page -- it looks nice -- and thought I'd try the puzzle. If you start both hourglasses simultaneously at time 0 and flip each one promptly when it finishes, you can measure off any number of minutes between 0 and 10 (and beyond; I didn't enumerate them past 10). The time line below shows when each hourglass gets flipped. Using time 0 as your initial start time, compute the elapsed time between the current start time and each flip of either hour glass, until either you find an elapsed time of 9 minutes (the solution) or you find an elapsed time greater than 9 minutes, in which case pick the next higher available start time and try again. Counting time 0 as a flip then it is 9 minutes from the second flip of the 7- minute hourglass at time 7 to the fifth flip of the 4-minute hourglass at time 16. The total time it took to do this was 16 minutes, which should be minimal unless there's some trick to it I didn't see. Time Flips Start time End time Elapsed time 0 4,7 0 0 0 1 | 0 4 4 2 | 0 7 7 3 | 0 8 8 4 4 0 12 12 (Bigger than 9 -- start over) 5 | 4 4 0 6 | 4 7 3 7 7 4 8 4 8 4 4 12 8 9 | 4 14 10 (Bigger than 9 -- start over) 10 | 7 7 0 11 | 7 8 1 12 4 7 12 5 13 | 7 14 7 14 7 7 16 9 (Bingo!) 15 | 16 4 17 | 18 | 19 | 20 4 21 7 22 | 23 | 24 4 25 | 26 | 27 | 28 4,7 From: Karl Hahn First, there is no "quickest way" to measure nine minute. By definition, if you measure nine minutes, it will take you exactly nine minutes. There is no quicker or slower way than that. Now to the solution. I had hoped there would be a trick, like having to lay one of the glasses on its side for some period, or something like that, but no -- the answer is very straightforward. Assume both glasses have start with all the sand in the bottom. 1) At time of 0 minutes, invert both glasses. 2) At time of 4 minutes the 4 minute glass has run out. The 7 minute glass still has 3 minutes left to run. Invert the 4 minute glass. 3) At time of 7 minutes the 7 minute glass runs out. The 4 minute glass still has 1 minute to run. Invert the 7 minute glass. 4) At time of 8 minutes, the 4 minute glass runs out again. The 7 minute glass has 6 minutes to run, but has exactly 1 minute of sand accumulated in the bottom. Invert the 7 minute glass. 5) At time of 9 minutes, the 7 minute glass runs out. Now, if what you meant by the "quickest way" is, in fact, the method with the fewest inversions, then logic tells me that my solution is that one. Reasoning is as follows. At each step you have a choice of inverting either of the glasses, both of the glasses, or doing nothing, for a total of 4 choices. Here is the decision tree. If at step 1 I invert only the 7 minute glass, then at 7 minutes I am left with a 4 minute timer and a 7 minute timer, giving me totals of 11 or 14 minutes. So that is out. If at step 1 I only invert the 4 minute glass, then at 4 minutes I can invert either or both of the glasses. If I invert only the 7 minute glass, nothing happens until 11 minutes, so that is out. If I invert only the 4 minute glass, then at 8 minutes I have a 4 minute timer and a 7 minute timer, for totals of 12 and 15 minutes, so that's out. That leaves only inverting both at step 1. If I do that, then when the 4 minute glass is done, I can: Do nothing. No measurement is them made after 4 minutes, so that's no good. Invert the 7 minute glass only. It will take 4 minutes to run out, so at 8 minutes I will be left with a 4 minute timer and a 7 minute timer for totals again of 12 and 15 minutes. No good. Or at step 2 I can invert the both glasses. Same result. Or at step 2 I can invert the 4 minute glass only. By process of elimination, that is what I must do. The next glass to run out will be the 7 minute glass at 7 minutes. At that point (step 3) I can: Do nothing. The 4 minute glass runs out at 8 minutes, leaving me again with a 4 minute timer and a 7 minute timer for totals of 12 and 15 minutes. No good. Invert the only the 4 minute glass. It will then run out at 10 minutes, which is no good. Invert both glasses. Again, nothing will happen until 10 minutes, which is no good. By elimination again we have that we must invert only the 7 minute glass. The next thing to happen will be that the 4 minute glass runs out at 8 minutes. At that point, there is 1 minute of sand in the bottom of the 7 minute glass, and the path to the solution is obvious. Since at every point I eliminated all possibilbities except the one I ended up taking, the path I took must therefore be the path of minimal inversions. Take care, Karl From: "William R. Wagenseller ;-{>" Organization: Heald Colleges Fastest time: 9 minutes Number of operations: 4 (1) Turn over both timers. (2) When the 4 min timer runs out, turn it over - TIME SINCE LAST ACTION = 4 MINUTES (3) When the 7 min timer runs out, turn it over - TIME SINCE LAST ACTION = 3 MINUTES - TOTAL ELAPSE TIME = 7 MINUTES (4) When the 4 min timer runs out, turn the 7 MINUTE TIMER over - TIME SINCE LAST ACTION = 1 MINUTE - TOTAL ELAPSE TIME = 8 MINUTES ... When the 7 min timer runs out, PROBLEM SOLVED... - TIME SINCE LAST ACTION = 1 MINUTE - TOTAL ELAPSE TIME = 9 MINUTES From: eosm@aqua.civag.unimelb.EDU.AU (Emma Osman) Nine minutes Start both off at the same time. At four minutes, turn the four-minute one over. At seven minutes, turn the seven-minute one over. At eight minutes, the seven minute one has been going for one minute. Stop it and turn it over, to re-measure the one minute. At the end of that, you are at nine minutes. Voila! I acheived this result by being morally certain that the answer was going to turn out to be "nine", then adding and subbing fours and sevens till I got it Emma Osman From: John Orthwein Let T be the set of whole numbers, representing time. At t=0, both hourglasses are turned over. A t=4, the 4-minute hourglass is empty and there is three minute's worth of sand in the 7-minute hourglass. At this instant, the 4-minute hourglass is turned over and started again. At t=7, the 4-minute hourglass now has one minute's worth of sand left. At t=8, one minute's worth of sand is in the bottom of the 7-minute hourglass and the 4-minute hourglass is empty. At t=8, turn the 7-minute hourglass over. When the sand runs out of the 7-minute's hourglass, t=9. One could now even measure time to any discreet minute by exploiting the fact that one hourglass is empty and the other has one minute's worth of sand in the bottom. From: THHACKETT@vassar.edu The way I would solve this problem is as follows: Start both hour-glasses at the same time. As each one completes, immediately re-start it. When the longer (seven-minute) hour glass completes for the first time, start the nine-minute interval. At that point, the four-minute hour-glass has one minute left on its second iteration. When the four-minute glass completes two more complete iterations, the nine-minute interval will have completed (sixteen minutes from the start of the whole thing). In order to arrive at this solution, I simply laid-out the seven and four minute intervals on a number line and found the first instance of a difference of nine. Intuitively I feel this is the shortest nine-mintue measurement, but I'm not yet sure how to prove it. Tom From: Michael Feld Organization: Philosophy Dept., University of Manitoba 1) The easy, obvious answer is: Invert the 4-minute sandclock; when it is half-run, invert the 7- minute sandclock; or reverse the order. Of course, this assumes keen eyes and a symmetrical sandclock. 2) The classical wise-guy answer is: "Oh mister time-keeping person, would you like to trade your cheap electronic time-keeping machine for this nice pair of sand-clocks?" 3) My stab at a real answer: Run both sand-clocks simultaneously,thus: 7 4 - - 0 0 at the end of four minutes, the clocks will look like this: 3 0 - - 4 4 Invert the smaller, i.e., the 4-minute glass, while letting the larger, i.e., the 7-minute glass, continue to run: thus, 3 4 - - 4 0 At the end of a further three minutes, the glasses will look like this: 0 1 - - 7 3 Next, invert the larger, 7-minute glass, while permitting the smaller, 4- minute glass to continue to flow. Thus, 7 1 - - 0 3 At the end of one minute, the glasses will appear: 1 0 - - 6 4 Invert only the larger, 7-minute glass, *AND START YOUR TIMING*. At the end of one minute, the glasses will appear: 0 0 - - 7 4 Next, invert only the smaller, 4-minute glass, *AND CONTINUE THE TIMING*; then, fo course, invert the 4-minute glass again. The last nine minutes of the process give you what you want. To reach those last nine minutes, you have used an additional eight minutes. So: the answer I reach is: 17 minutes in all Best, Michael Feld -- --------------------- From: Mike Knight Subject: Hourglass puzzle Excellent page and puzzle. Here is my answer. Solution: Label the 4-minute hourglass 'A', and the seven-minute glass 'B'. Start with both glasses, A and B, sand in the bottom. Flip both. Allow A to drain completely. At this time, 4 minutes have gone by. 3 minutes remain, however, in the B hourglass. Flip the A glass only. Allow the B hourglass to finish draining. 7 minutes have elapsed. One minute remains in the A glass. Flip the B hourglass. When A has drained, the 8 minute mark, flip B. Allow B to drain, measuring the last minute. Unfortunately, I can't provide a competent explaination of why this works. I simply know that we must find some way of splitting the total time into increments that we can measure with our hourglasses. The 4+3+1+1=9 was the idea that worked for me, other than trying to tell when half the sand had run out in the 4-minute hourglass, then starting the 7-minute glass. Also, in the real world, reaction time will play a part in our going slightly over the nine-minute mark. But, in puzzledom, human error need not apply. From: George Drapeau Hello, I've never visited this page before (Zeno's Coffeehouse) but came to it via a research page at Cornell on Project Zeno. Anyway, if the puzzle is serious, then here is my answer: 0:00 Start both hourglasses, the 4-minute and 7-minute glass. 4:00 When 4 minutes have elapsed, notice that the larger glass will have 3 minutes' worth of sand in it. Immediately turn over the now-empty 4-minute hourglass and let the larger one continue to drain. 7:00 The large hourglass has now drained and the small glass has drained 3 minutes' worth, leaving one minute left. Immediately turn over the larger hourglass, let the small one drain. 8:00 The small hourglass has drained a second time. Toss it out; you no longer need it. Notice that the larger hourglass has drained exactly one minute. Immediately turn it over and drain that minute. 9:00 Problem solved! Time for coffee. From: scott haney Well, the *quickest* way would take exactly 9 minutes. :) I don't have time to check if this is the shortest, but it should work: start both hourglasses restart the 4-minute glass when it runs out restart the 7-minute glass when it runs out when the 4-minute glass runs out, restart the 7-minute glass immediately! It should run for 1 minute. When it runs out, 9 minutes have passed. -- Scott Haney rhaney@cacd.rockwell.com (319)395-8281 Received: from ballard.staff.umkc.edu by axp2.umkc.edu (MX V4.1 AXP) with SMTP; Wed, 12 Jun 1996 14:56:57 CST I'm not sure that the challenge is to outfox a paradox here, but it seems to me that the fastest way to time nine minutes with a four minute hourglass and a seven minute hourglass, asuming perfect attention and instantaneous flipping, starting from zeroed hourglasses, is to 0 minutes 1: flip 'em both 4 minutes 2: flip the empty four minute hourglass, to time to 8 minutes. 7 minutes 3: flip the now empty seven minute hourglass 8 minutes 4: flip the seven minute hourglass that has just accumulated one minute of sand, which takes us to 9 minutes. <~\_/~\_/~\_/~\_/~\_/~\_/~\_/~\_/~\_/~\_/~\_/~\_/~\_/~\_/~\_/~\_/~\_/~> David_Nicol dnicol@cctr.umkc.edu pob45163_kcmo64171 (816)_235_1187 "beware of the leopard" From: mkeef@ix.netcom.com (MARV KEEFER) Subject: Timimg Puzzle Start both hourglasses. Turn over the four minute glass (four minutes.) Turn over the seven minute glass (seven minutes.) Turn over the seven minute glass when the four minute glass finishes (eight minutes.) Nine minutes have passed when the sand of the seven minute glass finishes draining. From: Ken Duisenberg Nine minutes can be measured in nine minutes using a four-minute and seven-minute hourglass: Start both hourglasses 0 minutes Restart the 4-minute glass 4 minutes Restart the 7-minute glass 7 minutes When the 4-minute glass runs out, turn over the 7-minute glass (it ran one minute) 8 minutes When the 7-minute glass runs out, the timing is done. 9 minutes Ken Duisenberg Roseville, CA From: shack@esinet.net (Shack Toms) Assuming that both timers start with all of the sand on one side, it can be done in 13 minutes. Otherwise I may have to wait up to an additional 3 and a half minutes to get the timers ready to start. The deductive principle is this, an hourglass can measure any interval of time and then duplicate that interval by letting the sand flow in reverse. So if I can get one timer in an initial state (all sand on one side) and a known time interval on one side of the other timer, I can duplicate that known time interval again and again, with one hourglass measuring the known interval while the other hourglass returns to its initial state. To be more explicit, at t=0 I start both timers. When the 4 minute timer runs out (at t=4) 3 minutes of sand remains in the seven minute timer. Now I can apply the above principle to measure three multiples of 3 minutes. I invert the 4 minute timer, to measure the 3 minutes remaining in the 7 minute timer. At t=7, the seven minute timer has emptied. I invert both timers so that with the 7 minute timer I can measure the 3 minute interval "stored" in the 4 minute timer. At t=10, the 4 minute timer is back in its initial state and 3 minutes of sand has drained from the 7 minute timer. I invert the 7 minute timer. At t=13 (at the end of my 9 minute interval that began at t=4) the 3 minutes of sand has now drained back out of the 7 minute timer. Shack Toms shack@esinet.net From: "Antonio Giosue'" First, start both hourglasses. When the 4m. hourglass ends, start it again.(4 minutes) When the 7m. hourglass ends, start it again.(7 minutes) When the 4m. hourglass ends, turn the 7 m. hourglass.(8 minutes) When the tm. hourglass ends we have measured nine minutes.(9 minutes) From: Mark Young Organization: Acadia University 9+7 = 16 = 4*4, so... T-7 minutes: Start both timers T-3 minutes: Flip the 4 (it just ran out) T: Begin measured nine minutes (the 7 just ran out) T+1: Flip the 4 (it just ran out) T+5: Flip the 4 (it just ran out) T+9: End measured 9 minutes (the 4 just ran out) ...mark young From: Michael Faron If the 9 minutes to be measured needs to be contiguous, then the "quickest" way i have found needs 7 minutes to get to the 9 minute interval.... at the same time, start the 4 and 7 minute hourglass when the 4 has emptied, immediately invert it. the 7 continues with 3 minutes left. when the 7 finishes its last 3 minutes, there is one minute remaining in the 4. immediately flip the 4. this one minute begins the 9 minute interval. follow the one minute with two 4 minute hourglass intervals. thats 9 minutes. if the 9 minute interval can be broken up, the "quickest" way i have found needs 5 minutes in addition to the 9 minutes desired.... at the same time, start the 4 and 7 minute hourglass. when the 4 has emptied, immediately invert it. the 3 minutes left in the 7 begin the 9 minute interval. when the 3 minutes has finished, immediately flip both hourglasses. 1 minute is left in the 4 and begins to empty. the full 7 is emptying as well. when the 1 minute has emptied, there are 6 minutes left left in the 7 minute hour glass add these 6 onto the previous three and you get the desired 9. thats all i can come up with mfaron@sinet.com From: "Peter W. Woodruff" Dear Ron: Lovely site! Let t be the time; it must either be nine minutes since some time we turned the four-minute glass or since some time we turned the seven minute glass, and it must be marked as the expiration of the other glass. Put another way, the beginning of the interval must be marked by one glass, the end by the other. Thus we need either that the end is both 4f + 9 and 7s or both 7s+9 and 4f for some f and s. Calculation (or experimentation) yields that the minimum values for f and s in the first case are 3 and 3, in the second case 1 and 4; the first yields a time of 21, the second of 16, so 16 minutes is the shortest (of course, this assumes that we can start the process we want to measure at any time; if the start time is fixed, there is no solution). --Peter From: Francis Varela Using standard hourglass-type measuring devices, what is the quickest way to measure nine minutes with a four-minute glass and a seven-minute glass? Solution: Time 0 Step 1. Set both the 4-min. and the 7-min. glasses. After 4 min. Step 2. Invert the 4-min. glass. After 3 min. Step 3. The 7-min. glass is used up; start counting. After 1 min. Step 4. Invert the 4-min. glass. After 4 min. Step 5. Invert the 4-min. glass. After 4 min. Step 6. The 4-min. glass is used up; 9 min. have been counted. The 9-minute period is counted starting with Step 3. The whole procedure requires 16 minutes to perform. Francis Varela From: Roy Kilgard Come now, Barnette, this one's easy!!! To time nine minutes with a four minute and seven minute timer: Start both timers simultaneously. When the four minute runs out, start it again. When the seven minute runs out, begin timing nine minute interval. Let the remaining minute fall through the four minute and run it twice more. 7-4=3 4-3=1 1+4+4=9 Simple enough. -Roy From: JohnD_Halter The best way to time 9 minutes takes 16. YOu will have to flip the 4- minute glass so that it may consume the sand four times. The 7-minute glass will be flipped only to begin the process, and the timing of the 9 minute span should begin when this glass runs out. Thanks for the challenge. I'll be looking forward to the next one! JOhn From: Terrence Nolley I don't think it really makes a difference what combination of glasses you use. The process will still take nine minutes regardless of the combination. From: "Robert N. Kelly" Empty the four minute hourglass. Fill the four minute hourglass with the seven minute hourglass giving yourself three minutes left in the Seven minute hourglass. Turn the seven minute hourglass over three times for nine minutes. If this is illegal then Turn both hourglasses over at the same time. At the end of four minutes turn the four minute hourglass back over. At the end of the seven minute hourglass turn the four and the Seven minute hourglass back over. {one minute is left in the four minute hourglass.} At the end of that minute turn the seven minute hourglass over again. (one minute is left in the seven minute hourglass} Since the seven minute hourglass expired fully once, and you used the minute left in the four minute hourglass on the third turn over and you measured a minute of sand in the seven minute hour glass and turned that over you have measured NINE minutes. Seven plus one plus one. From: monlyn Subject: 9 MINUTES TI think the best way would be to do 1, 7-minute, then = of the 4-minute hourglasses.There is no "quicker" way, for 9 minutes is nine minutes! Just MY opinion,only........Monica From: Ed Russell Deduction: 1. The result cannot be obtained without measuring a time interval of 1, 2, or 5 by subtraction. 2. The subtraction process must be a continuous use of both the 4 and 7 glasses. 3. This leads to the differences in the series 7, 14, ... and 4, 8, 12, ... => (7-4=3), (7-8=-1).... Therefore of the time intervals specified in (1), 1 is the first to be measured. 4. The total time for measuring 9 minutes: 8 minutes to measure 1 minute, plus 2 times 4 minutes = 16. -- Ed Russell ehr@synchronicity.com From: Chad Miller Start the four- and seven-minute glasses at the same time. When the four- minute glass runs out, flip it. When the seven-minute glass ends, start timing; there is one minute left in the four-minute glass. When it finishes, flip the four-minute glass twice, adding eight more minutes. Viola! Chad, ex-phicyberite http://www.surfsouth.com/~cmiller/ 0400 Date: Mon, 1 Jul 1996 10:46:47 -0400 (EDT) From: Rex DeVane There is no quick way to measure nine minutes - it will take nine minutes any way you attempt to measure it. The following is one method using the seven and four minute hourglass type measuring devices: 1. Turn both the seven and the four minute glasses over to start them running 2. As soon as the four minute glass is empty turn it over to start a new four minutes. Note that the seven minute glass is still running and has only 3 minutes left. 3. When the seven minute glass runs out turn it over to start a new seven minutes. Note that when we you turn the seven minute glass over you have only one minute left in the four minute glass. 4. When the four minute glass runs out turn the seven minute glass back over. Note that the seven minute glass only had time to drop 1 minute of sand. 5. When the seven minute glass is empty you have measured nine minutes! From: Rex DeVane Subject: Charles' time problem To: rbarnett@grits.valdosta.peachnet.edu Message-ID: MIME-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Content-Length: 883 Status: O X-Status: There is no quick way to measure nine minutes - it will take nine minutes any way you attempt to measure it. The following is one method using the seven and four minute hourglass type measuring devices: 1. Turn both the seven and the four minute glasses over to start them running 2. As soon as the four minute glass is empty turn it over to start a new four minutes. Note that the seven minute glass is still running and has only 3 minutes left. 3. When the seven minute glass runs out turn it over to start a new seven minutes. Note that when we you turn the seven minute glass over you have only one minute left in the four minute glass. 4. When the four minute glass runs out turn the seven minute glass back over. Note that the seven minute glass only had time to drop 1 minute of sand. 5. When the seven minute glass is empty you have measured nine minutes! From: Math/CS Lab Turn both hour glasses over and let the 4 min. one run out. Stop the 7 min one as soon as the 4 min runs out. You have 3 min left on the 7 min one and 0 min on th e4 min one. Restart both hour glasses and let the 7 min one run out. You have 0 min on the 7 min one and 1 min on the 4 min one. Restart both hourglasses again. You'll have 6 min on the 7 min glass and 3 min on the 4 min glass. Start the 7 min glass and when it runs out start the 4 min glass. You get 9 minutes! Laura Received: from reid.phil.indiana.edu (reid.phil.indiana.edu From: Karen Brown Turn both glasses over. When the four minute one runs down, turn it over again. When the seven minute one runs out, the nine minutes begins. Allow the four minute glass to run out three times [1 minute + 4 minutes + 4 minutes]. From: "Christoffer J. Anderson" This is actually a pretty easy one. Here goes: (1) Start both timers going simultaneously. (2) When the Four-Minute-Timer (4MT) is done, flip it over and leave the Seven-Minute-Timer (7MT) to keep going its last 3 minutes. At this point, we have elapsed 4 minutes, and there is 3 minutes left on the 7MT. (3) When the 7MT is done, flip it over. At this point, we have elapsed 7 minutes, and there is 1 minute left on the 4MT. (4) When the 4MT runs out, there will have accrued 1 minute's worth of sand in the bottom of the 7MT. Flip the 7MT over when the 4MT is done. At this point, we have elapsed 8 minutes, and there is 1 minute left in the 7MT. (5) When the 7MT has run its 1 minute's worth of sand out, we have elapsed exactly 9 minutes. Gimme another! Christoffer J. Anderson Colorado State University Department of Philosophy Received: from Iris (193.219.56.3) by mbiserv.fermentas.lt 0. both glasses begin. 4. revert 4-glass. in 7-glass 3 min remaining. 7. 7-glass end, revert 4-glass with 1 min. remaining. start count. 8. revert 4-glass. 12. revert 4-glass. 16. end count. is 16 min the limit? -- v. tiknius vt@fermentas.lt http://iris.fermentas.lt From: SBIJ101@aol.com 1)Turn over both timers 2)When the four minute timer runs out turn it back over while the remaining three minutes runs out on the other timer. TOTAL 4min 3)When the seven minute timer is empty turn it back over while the remaining minute runs out on the four minute timer TOTAL 7min 4)When the four minute timer is empty turn the seven minute timer back over TOTAL 8min 5)When the seven minute timer is empty you have measured ----------- TOTAL 9min From: Gregory Reihman > Using standard hourglass-type measuring devices, what is the > quickest way to measure nine minutes with a four-minute glass and a > seven-minute glass? One can measure eight minutes by using the four-minute glass twice in succession. Conveniently, this is one minute longer than the time measured by the seven-minute timer, and one minute shorter than the total desired time. So, the desired nine minutes can be measured by first beginning the clocks at the same time, then flipping the four-minute clock when it runs out, and then flipping the seven-minute clock when it (the seven-minute clock) is finished. Finally, when the four-minute clock runs out a second time (at this point one minute's worth of sand will have fallen in the seven-minute clock), flip the seven minute clock. When this minute's worth of sand falls, a total of nine minutes will have elapsed. This is assuming that the sand falls at a constant rate throughout the seven-minutes of time (or, at least, that the first and last minute's worth of sand are the same). If, for example, the sand were to fall faster near the beginning (when most of the sand is still on top) than near the end (when most of the sand is at the bottom), then it will take more than a minute for the "minute's worth of sand" to fall in the final step, and the time elapsed in the above procedure will be more than nine minutes. If the sand does fall at a constant rate, the procedure will work. From: Christopher Hitchcock Re: New Zeno challenge You can time nine minutes in nine minutes. Start both hourglasses at minute zero. When the four minute glass is finished, turn it over. When the seven minute glass is finished, turn it over again. When the four minute glass finishes a second time (and eight minutes have elapsed) there will be one minute's worth of sand in the bottom of the seven minute glass. Turn it over, and it will take one more minute for the sand to fall through. Voila--nine minutes. From: MattSwarm@aol.com A fast way to measure nine minutes with a 4 minute and a 7 minute hour-glass type device takes 9 minutes. There are very few ways to do it faster. Minute 00: Start both 4 and 7 minute timer. Minute 04: Turn over empty 4 minute timer. Minute 07: Turn over empty 7 minute timer. Minute 08: 4 Minute timer has finished, 7 minute timer has 1 minute of sand in bottom. Turn over 7 minute timer. Minute 09: Aforementioned 1 minute of sand has drained to bottom. Question: What integer periods, if any, CANNOT be measured using this and similar methods? Matt Swarm mattswarm@aol.com From: Michael Britton simple enough, although it's eaven simpler to be trapped in the 3-minute pit. (i.e., trying to use the 7-4=3 and 3x3=9 properties... I'm sure some respondent will provide this solution in full. it requres 4 minutes preparation.) I'll use a wierd notation here: t=__ is the time from start; | x| >< | y| is an hourglass with x minutes of sand in the top and y minutes of sand in the bottom. the first hourglass will be the 4-minute one, and the second will be the 7-minute one. so initially, we start both hourglasses running: t=0 | 4| | 7| >< >< | 0| | 0| 4 minutes later, the first hourglass runs out, so flip it over: t=4 | 0| | 3| ---\ | 4| | 3| >< >< > >< >< | 4| | 4| ---/ | 0| | 4| 3 minutes later, the other hourglass runs out, so flip it over: t=7 | 1| | 0| ---\ | 1| | 7| >< >< > >< >< | 3| | 7| ---/ | 3| | 0| 1 minute after that, the first hourglass runs out again. this time, flip the second hourglasses over again. t=8 | 0| | 6| ---\ | 0| | 1| >< >< > >< >< | 4| | 1| ---/ | 4| | 6| 1 minute later, the second hourglass runs out of sand again. when this happens, 9 minutes have expired. t=9 | 0| | 0| >< >< | 4| | 7| and there you go! Mike Britton From: ryanearehart@juno.com To: RBARNETT@GRITS.VALDOSTA.PEACHNET.EDU IINTERESTING QUESTION MY SOLUTION TAKES 12 MINUTES TO SET UP BEFORE YOU CAN MEASURE YOUR NINE MINUTES. fIRST YOU START THEM BOTH GOING AT THE SAME TIME AND WHEN THE 4 MINUTES FINISHES KNOCK OVER THE SEVEN SO THAT IT HAS 3 MINUTES IN ONE HALF AND FOUR IN THE OTHER. USE THE 3 MINUTES TO REDUCE THE 4 MINUTE ONE TO ONE MINUTE AND 3 MINUTES - - THEN TAKE THE 1 MINUTE AND REDUCE THE SEVEN TO SIX AND ONE. FROM THE SIX DEDUCT FOUR LEAVING YOU WITH 2 MINUTES AND 5 MINUTES LEFT IN THE BIG ONE AND 4 MINUTES IN THE SMALLL ONE USE THE 4 AND THE FIVE TOGETHER AND I THINK THAT ADDS UP TO NINE! {4} - {7} {4}-{3/4} 4MIN {1/3}-{7} 3MIN {4}-{6/1} 1MIN {4}-{5/2} 4MIN ELAPSED TIME = 12MINUTES 4+5=9 Received: from xyp29p1.ltec.net (xyp29p1.ltec.net [204.96.104.11]) by iac1.ltec.net (8.6.10/8.6.10) with SMTP id UAA39373 for Ok, start the 7 minute glass and the four minute glass at the same time. When the four minute glass runs out, start it again. When the seven minute glass runs out, begin timing. When the four minute glass runs out, one minute will have elapsed. Start the one minute glass again. When it runs out, five minutes will have elapsed. When it runs out again, you will have measured nine minutes. Matt Seaman From: "Michael O'Leary" 1.Flip both 4 and 7 hourglasses over. 2.When the 4 glass is done, flip both 4 and 7 over (with 3 minutes left in 3. 3. When the 7 (with 3 min lifet in it is done, flip over the 4 (with one minute left) and the 7 glass. 4. When the 4 (with one minute left is done, flip the 7 over. It will have 1 minute left in it. That will take 9 minutes. -- _\|/_ O O +----------------oOO-(_)-OOo-----------------------+ Michael O'Leary From: Lauri Kultti Organization: University of Helsinki First turn both glasses. When the four-minutes glass is done, the seven-minutes glass has three to go. Turn the four-minutes glass again. When the seven minutes-glass (with its three minutes left) is done, the four-minutes glass has one to go. START THE MEASURING NOW. Let the remaining one minute (of the four-minutes glass) go. Then simply use the four-minutes glass twice. So, you have your nine minutes. From: Tony Reardon 16 mins elapsed. To make 9 minutes you either need a 2 minute interval to add to 7 or a 1 minute interval to add to 4 plus 4. 2 minutes cannot be derived until 14 minutes have elapsed but 1 minute can be derived after 7 minutes. There are a couple of ways of doing this but the simplest is to start both timers. After 4 minutes, restart the 4 minute timer. After 7 minutes restart the 4 minute timer which will run for 1 minute. Then use the 4 minute timer twice more to get 9 minutes. From: Wayne BARNES The process will take a total of 25 minutes. Begin by flipping both glasses. When one glass is finished note the amount of time remaining in the other. After 16 minutes there will be 5 minutes remaining in the 7 minute hour-glass. You can now begin measuring the required 9 minutes by turning the 4 minute hour-glass after this five minutes has expired. From: Dave Evans Hi, The quickest way I can figure is to have one 4 minute hourglass and three 7 minute hourglasses. Turn them all over at once, when the 4 minute glass is done, turn 2 of the seven minute glasses on their sides, so they stop flowing. The nine minutes starts now. When the 1st seven minute glass is done, restart the 2nd, when the second is done, restart the 3rd. Since each had 3 minutes left (7-4=3), when the third is done, 9 minutes has elapsed. Grand total of 13 minutes to measure 9.