35
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You have two hourglass timers. One can measure 7 minutes and the other hourglass timer can measure 13 minutes. This means that it takes 7 minutes for the sand timer to completely empty the sand from one portion to the other.

Your task is to measure 22 minutes using both the timers in the shortest time after the task begins, that means your time starts after you start to play with the hourglasses until you can measure real continuous 22 minutes.

So how long at the earliest will it take to measure 22 minutes after your task begins?

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  • 1
    $\begingroup$ This puzzle seems familiar to me. Did you create it, or did you find it somewhere? $\endgroup$ – Challenger5 Aug 12 '17 at 20:29
  • $\begingroup$ @Challenger5 22 part is my original, hourglass is classical puzzle. $\endgroup$ – Oray Aug 12 '17 at 20:41
  • $\begingroup$ Okay. The numbers seem familiar for some reason. $\endgroup$ – Challenger5 Aug 12 '17 at 20:54
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    $\begingroup$ Is laying 1 of the hourglasses flat so there is no throughput for some time allowed? $\endgroup$ – Zaibis Aug 14 '17 at 11:42
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    $\begingroup$ As I wrote earlier, I had the following (physics-related) doubt. If I start a sand glass timer, let it go for a minute and then flip it, will it go exactly for a minute? It turned out this is a reasonable assumption, as sand flow is much different from fluid flow (inf.utfsm.cl/~amoreira/orzelc.pdf , Eur. J. Phys. 17 (1996) 97–109). "The flow was constant at 0.67 g s−1 throughout most of a 19 m 45 s emptying period, only dropping to 0.62 g s−1 during the last minute." I experimented with a 10 min sand glass. I ran the sand glass for 46 s, then flipped it, then it ran for 45 s. $\endgroup$ – akhmeteli Aug 15 '17 at 3:20
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A very simple solution, taking

35 minutes

is to

start ($T+0$) with turning both hourglasses, and keep turning the 7-minutes one. The period of 22 minutes starts after the 13-minutes hourglass finishes ($T+13$); after the 7-minutes one completes 5 runs, it will be $T+35$, 22 minutes after $T+13$.

A better solution takes only

22 minutes, so it's an optimal solution.

How?

Start ($T+0$) with only the 7-minute hourglass. After 7 minutes ($T+7$), start both of them. After another 7 minutes, turn only the 7-minute one and keep the 13-minute one running. When that one finishes ($T+20$), there's still one minute left in the 7-minute one. Flip the 13-minute one, it will run for one minute until the 7-minute one finishes ($T+21$). Now, flip the 13-minute one again, and it will run for 1 minute, finishing at $T+22$.

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  • $\begingroup$ well done! :) very smart of you! $\endgroup$ – Oray Aug 12 '17 at 10:58
  • $\begingroup$ Thanks! I was wondering why you already accepted my answer before I posted the second solution ... or does that use some loophole which isn't allowed? $\endgroup$ – Glorfindel Aug 12 '17 at 11:00
  • $\begingroup$ it was accidental :) and was going to take back then noticed you edited it with my intentional answer, so i did not take it back :) $\endgroup$ – Oray Aug 12 '17 at 11:01
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    $\begingroup$ That second solution only works if you assume that the amount of sand is linear with respect to time, whereas I assumed you could only measure the whole interval with the hourglass. $\endgroup$ – someonewithpc Aug 14 '17 at 14:26
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    $\begingroup$ @someonewithpc true, but this is Puzzling, where we make assumptions because they make the puzzle interesting. $\endgroup$ – Glorfindel Aug 14 '17 at 16:52
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22 minutes with lateral thinking (though the accepted answer is better)

Start both timers, and when the 7 runs out, flip it again. When the 13 ends, flip it again. When the 7 runs out, the 13 has one minute at its bottom. Put the 13 on its side (stop the sand), and flip the 7. The 7 minute runs out at +21, and the remaining minute is in one side of the 13.

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    $\begingroup$ Oh, sh$%, you and I had the same idea. $\endgroup$ – user39732 Aug 13 '17 at 5:57
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    $\begingroup$ Did you mean to make that pun? Lateral thinking... laying it on it side? $\endgroup$ – josh Aug 14 '17 at 15:18
  • $\begingroup$ This is better than the accepted answer as you do not need an additional timepiece. +1 $\endgroup$ – Joe Aug 14 '17 at 17:58
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    $\begingroup$ @Joe extra timepiece? Both answers take advantage of 14 minutes being one greater than 13. Glorfindel's answer gets that minute when it's needed, while I had to find a sneaky way to save it for later. $\endgroup$ – Carl Aug 14 '17 at 19:19
  • $\begingroup$ I like the idea, although we don't know if the hourglasses can be tipped sideways without sand running anywhere. Maybe they are shaped so the sand always equalizes if laid on the side. $\endgroup$ – mathreadler Aug 15 '17 at 8:54
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Here's an approach that will let you measure any length of time (including 22 minutes), to the nearest minute, after a few initial moves that take 13 minutes total:

Notation: Let the state of an hourglass H be represented by H(x,y), where x is the amount of sand on top, y is the amount of sand on the bottom, and h=x+y equals the total length of the hourglass (i.e. 7 or 13 minutes).

The approach is to set up hourglasses A and B such that the first has state A(a-1,1) and the second has state B(0,b), where a=7 and b=13 are the total volumes of hourglasses A and B. Once you get to this state, you can simply flip both hourglasses every minute - the first hourglass will empty after one minute and the other will be left with one minute of sand remaining. You can then flip both glasses every minute for as many minutes you need to count.

Now to get into this position:

Start at t=0 with A(7,0) & B(13,0). At t=7min they have drained to A(0,7) & B(6,7). Now flip A, so you have A(7,0) & B(6,7). After six minutes, at t=13min, you have A(1,6) & B(0,13). Flip A, and we are now in the position where we can count every additional minute up to 22 minutes or more.

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H1 = 7m hour glass, H2 = 13m hour glass. H1 sand and H2 sand at the bottoms of both glasses, flip both at the same time. When H1 runs out flip it again quickly leaving H2.

AT THIS POINT, 7 MINUTES HAS GONE BY.

H2 HAS 6 MINUTES IN ITS TOP.

H1 HAS 7 MINUTES IN ITS TOP.

When H2 runs out, flip H2 again leaving H1.

AT THIS POINT, 13 MINUTES HAS GONE BY.

H2 HAS 13 MINUTES IN ITS TOP.

H1 HAS 1 MINUTE IN ITS TOP.

When H1 runs out, set H2 with its top-to-bottom axis parallel to the table, and flip H1 again, at the same time, storing H2's 1 minute in the "bottom," now sideways.

AT THIS POINT, 14 MINUTES HAS GONE BY.

1 MINUTE IS STORED SIDEWAYS IN H2, WHILE

H1 HAS 7 MINUTES IN ITS TOP.

When H1 runs out, set H2 to have its 1 minute in the top position. When H2 runs out, 22 minutes will have gone by.


Below, the gray side is the 7 minute hour glass, and the white side is the 13 minute hour glass. Anytime consecutive squares have the same hour glass configuration indicates the other hourglass has been flipped. This shows the process visually:

enter image description here

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  • 1
    $\begingroup$ (The explanation provided earlier by @Carl seems simpler to understand than this one, to me. In general, you should check other answers before posting your own, to avoid duplication.) $\endgroup$ – Rubio Aug 13 '17 at 7:18
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Start both 7 minute (HG1) and 13 minute (HG2) hourglasses at the same time.

After 7 minutes elapses, the 13M (HG2) hourglass will have 6 minutes remaining.

Flip the 7 minute hourglass (HG1) over again to start over.

After the HG2 finishes its 6 remaining minutes, there will be 1 minute remaining on the HG1, 7m hourglass.

That 1 minute remaining is the starting point.

1 minute, then 3* 7m hourglasses = 22 minutes.

You can start your task after 13 minutes.

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protected by Rand al'Thor Aug 14 '17 at 16:57

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