-4
$\begingroup$

A Kangaroo wants to cross a 3 lane highway, all cars going in same direction. All lanes are 7 feet wide.

In lane A a car passes every 4 seconds

In lane B a car passes every 3 seconds

In lane C a car passes every Prime Number seconds (2,3,5,7,11,13,17,19,23 etc seconds)

The Kangaroo can only do the following:

Jump forward exactly 5 feet in 3 seconds. Once he jumps cars cannot hit him. So it is instantaneous.

Jump backwards exactly 2 feet in 3 seconds

Wait

He can ONLY go in straightline perpandicular to the lane direction. No fractional seconds in this puzzle. The (zero) time starts when all cars pass simultaneously in front of him so the next car in lane A will pass in 4 seconds, one in lane B will pass in 3 seconds and one in lane C will pass in 2 seconds. The next car in lane C will pass at 3 second mark the next one at 5 second mark, the next one at 7 sec mark and so on.

Mr. Kangaroo needs to get across the 3 lanes in less than 25 seconds.

Can he do it? What do you suggest? enter image description here

$\endgroup$
  • $\begingroup$ I assume we are supposed to believe the drivers won't notice or care about a kangaroo, meaning they do not alter their driving when they see it on the pavement ahead? $\endgroup$ – Ian MacDonald May 1 '17 at 13:40
  • $\begingroup$ going forward at a pace of 1.67ft/s, wont the kangaroo get run over in lane A no matter what since he can only travel 6.67 ft in 4 seconds? or is it that he jumps instantaneously forwards/backwards at the end of 3 seconds (and therefore is at his originally position until the mark of his 3 second jump is completed)? $\endgroup$ – indubitablee May 1 '17 at 13:57
  • $\begingroup$ Good question Please assume that while jumping he does not get hit. Second comment by indubitablee correct. Instantaneous jump $\endgroup$ – DEEM May 1 '17 at 14:03
  • $\begingroup$ Does "in less than 25 seconds" mean (1) less than 25s after the start time, or (2) less than 25s after the start of the kangaroo's first jump? $\endgroup$ – Gareth McCaughan May 1 '17 at 14:21
  • 1
    $\begingroup$ There is an important distinction between "at his original position until 3 seconds elapse, and then the jump is instantaneous", and "the jump takes 3 seconds but is high enough in the air that no car will hit him" (so that he is in fact NOT in his original position for those 3 seconds). Which is intended? $\endgroup$ – Rubio May 1 '17 at 14:41
4
$\begingroup$

If I am understanding the problem correctly, the following seems like it works.

t=0. Start just south of the A lane. (Not in the road.)
Wait 4s
t=4. Just south of the A lane. (Not in the road.)
Jump forward
t=7. Almost 5ft into the A lane. (Not a multiple of 4s.)
Jump forward
t=10. Almost 10-7=3ft into the B lane. (Not a multiple of 3s.)
Wait 1s
t=11. Almost 10-7=3ft into the B lane. (Not a multiple of 3s.)
Jump forward
t=14. Almost 8-7=1ft into the C lane. (Not a prime time.)
Jump backward
t=17. Almost (-1)+7=6ft into the B lane. (Not a multiple of 3s.)
Jump forward
t=20. Almost 11-7=4ft into the C lane. (Not a prime time.)
Jump forward
t=23. Out of the C lane by 2ft. (Not in the road.)

But perhaps I am misunderstanding and

we aren't allowed to go so far past the far edge? Perhaps in particular we're meant to start exactly on the south edge of the A lane, in which case I would like to reiterate Sconibulus's question about exactly what that means.

If we interpret the question differently, so that you have to start and end exactly at the edge of the road, and if we suppose that somehow that doesn't imply being unavoidably hit by a car at t=0, then

you have to travel a total of 21ft forward in steps of +5 and -2. So your number of backward steps must be 2 (mod 5). We clearly don't have time for 7 or more backward steps, so there must be exactly two, so we need (21+4)/5 = 5 forward steps. That's 5 forward, 2 backward, for a total of 7 jumps taking up 21s of the available time. That may well be possible but it doesn't in fact sound as if this is what OP intends so I'll drop this line of thought now.

$\endgroup$
  • $\begingroup$ It takes three seconds to jumps, so I don't think you can act at t=10 and t=11 $\endgroup$ – Sconibulus May 1 '17 at 14:45
  • $\begingroup$ Sorry, the 11 was meant to be 13 and everything subsequent to it later by 2s. Edited now. $\endgroup$ – Gareth McCaughan May 1 '17 at 14:52
  • $\begingroup$ Gareth. I think you know my intention. There is an answer for <25 seconds. But technically you have it. $\endgroup$ – DEEM May 1 '17 at 14:55
  • $\begingroup$ OK, got one with <25s. $\endgroup$ – Gareth McCaughan May 1 '17 at 15:00
  • $\begingroup$ so this solution is on the premise that the Kangaroo is above the cars during the jump and is only on the ground the instant that he lands or when he waits. nice solution sir $\endgroup$ – indubitablee May 1 '17 at 15:11
1
$\begingroup$

If he doesn't wait at all, he stops every 3 seconds, getting already hit on Lane B. So he must wait for a while before getting there. Waiting on the side is more efficient in every way compared to jumping forward once and waiting on Lane A or going back to the beginning, meaning we can just assume that the latter two aren't done at all on the first lane.

Let's say he waits for $n$ seconds at start. Then this $n$ must not be written as $4k+1$, or he gets hit only 3 seconds later. Also, $n$ must be indivisible by 3 to prevent getting hit in Lane B (possible if $n = 12x+2,4,7,8,10,11$). Only $n=2,4,7,8$ allow him to cross the road in less than 25 seconds. When he does the second jump, he can stop at the 8th, 10th, 13th or 14th seconds.

Lane C can be crossed in two jumps, so ultimately he must start jumping a composite number of seconds ($c$) after the beginning at the 15 meter mark and $c+3$ must also be composite. The only possible $c$ values allowing a shorter trip than 25 seconds and that could be realistically reached are the multiples of 3 from 12 to 18 (and it gives a prime if $c$ isn't divisible by 3).

At one point, he jumps from B to C to the 20 meter mark. In this case, he can't be starting this jump a composite number of seconds after the beginning, otherwise he ends up at C at a prime-numbered second mark or can't even survive B in the first place. The prime in question (B to C) can't be smaller than 11 or bigger than 13. Before jumping from the 15 meter mark to the 20 meter mark, he must then wait until the soonest $3k$th second, which can only be done if he starts jumping from B to C at the 11th second.

Solution:

Wait for 4 seconds, jump forward twice, then wait for one second, later jump forward once, wait for one second again, and jump forward twice.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.