There is a line of bombs, all over a building. You are the bomb defusal expert, and you are the one everyone is counting on the save the city, for as if any one of the bombs go off, you, and everyone in the city, will die a tragic death. Obviously you don't want that to happen, so you read up the secret file they gave you on what's going on so you can get started.


You have to stall the bombs until we can hack into the entire system and shut all the bombs down. You have to run from button to button located on the bombs. Touching the buttons on them will reset the timers on them.

Here's a diagram of the bombs, how long it will take to get to them, etc.

A <==1m==>  B <==2m30s==> E

/\         /\
2m         4m
\/         \/

C <==3m==> D

A's timer is 10m
B's timer is 12m
C's timer is 12m
D's timer is 16m
E's timer is 12m

You are starting off at bomb A. Also, it does not take any time actually pushing the button on the bomb. The only time spent (Hopefully, if you're not just standing on the wall doing nothing, which you can do, but make sure you have enough time...) is actually running between the bombs. If you get to a bomb with no time left you can use your ninja skills to get the bomb's button pushed before it explodes, but if you're in the middle of a pathway, like if you were 2 minutes in on a 3 minute pathway, and the bomb you were travelling to went to 0, you couldn't suddenly teleport there to press the button.


You decide to get started on you're task, and you get a message that they've starting the hacking already.


Little does anyone know that in reality the hacking will fail every time. Just how long can you stall the bombs so that you get the maximum time before everything explodes?

Try to do this without computer programs, because that would be too easy!

P.S. Please no loopholes, there's no lateral-thinking tag here!


2 Answers 2


I think I got 19m30s...

(A)CABEBACD(E explodes)

  • $\begingroup$ I think this is correct, you arrive at D just in time, but cant make it to E, but it took you 7,5 to get to E, so 19,5.. $\endgroup$
    – DrunkWolf
    Nov 12, 2015 at 5:46
  • $\begingroup$ Ah, 2m + 11m (CABE round trip) + 3m = 16 to get to D. $\endgroup$
    – JonTheMon
    Nov 12, 2015 at 13:51
  • $\begingroup$ I haven't looked at Puzzling for a while, but yes this is a better one. It looks like I didn't notice this before! Gasps DUN DUN DUN! Seriously though, congrats! :D $\endgroup$ Nov 12, 2015 at 17:17

I'll say a guess of 18m30s.

C and E are you opposite points. While we could make a CABE circuit (11m), that leaves out D, which will eventually blow.

EBACD one-way takes 8m30s

So, ABE (3m30s), wait 3m, EBAC (5m30s, just enough to prevent C blowing), D (3m), wait 3m30s for E to blow.

  • $\begingroup$ I remember this being the correct answer, but if anyone else can get a better answer, I'll have to give it to them. I'm pretty sure, though, that that is the best answer there is. $\endgroup$ Nov 11, 2015 at 23:04
  • $\begingroup$ At the end you are waiting for E to blow at 3:30, not C. $\endgroup$
    – Paul
    Nov 11, 2015 at 23:07
  • $\begingroup$ Aha, yes Paul is correct, you might want to fix that. $\endgroup$ Nov 11, 2015 at 23:12
  • 2
    $\begingroup$ This is not the correct answer, you go back from e to get to c in time, but you have time to visit c before going to e, getting you an extra minute $\endgroup$
    – DrunkWolf
    Nov 12, 2015 at 5:54

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