Sohrab is taking a long walk. He always finds it interesting to see how many sidewalk cracks he stepped on. The cracks are 5 1/4 feet apart. The cracks are very thin. His stride is 2 3/4 feet long (measured from heel to heel). Assume his foot has to overlap a crack by at least an inch for him to count it. If he starts his walk with both heels just barely past a crack, what fraction of the cracks will he step on during his walk? (Beginning of the cycle starts when his heel is just past a crack. Cycle is the number of cracks. Only count the beginning or end crack, not both, just count one of them.)The fraction of the cracks he steps on is the number of cracks he steps on divided by the number of cracks in the cycle.

The distance will be the cycle from start to end. The cycle starts when his heels are past the first crack, and it will end when he reaches the crack on the sidewalk where his feet are in the same place as the starting position I am assuming?

  • $\begingroup$ So how long does he walk for? 1 mile? $\endgroup$ Commented Apr 8, 2014 at 15:21
  • $\begingroup$ That is one of the reasons I am really confused. It doesn't give a specified distance of how far he is walking. $\endgroup$
    – Samantha
    Commented Apr 8, 2014 at 16:33
  • $\begingroup$ How long are his feet? $\endgroup$
    – Jack M
    Commented Apr 8, 2014 at 21:24

1 Answer 1


crack distance = x = 5.25 feet
pace distance = y = 2.75 feet

We will start by finding how may strides and cracks are in a cycle (that is, how many steps are taken/how many cracks are passed before Sohrab is just past a crack his heel is touching). This is called the least common multiple (LCM) and is most easily found by simply listing all the multiples of X and Y and selecting the first number that appears twice. In our case, this number is 57.75.

This means that we have 57.75/5.25 or 11 cracks per cycle. Similarly we have 57.75/2.75or 21 steps per cycle.

We can check each crack to see if it falls within 11 inches of a foot step (because a crack must be overlapped by at least an inch, and I'm assuming his foot is a foot in length). 11 inches is 0.9166.. feet, so for any multiple of X < 11 and Y < 21 such that X < Y <= X+0.9166, we've located a hit. We can check this manually because the numbers are small. First, list the step distances:

2.75, 5.5, 8.25, 11., 13.75, 16.5, 19.25, 22., 24.75, 27.5, 30.25, 33., 35.75, 38.5, 41.25, 44., 46.75, 49.5, 52.25, 55., 57.75

(From this query)

Next list the crack distances:

5.25, 10.5, 15.75, 21., 26.25, 31.5, 36.75, 42., 47.25, 52.5, 57.75

(From this query)

Now check each crack distance to see if it satisfies our formula:

5.25 : yes (hits 5.5)  
10.5 : yes (hits 11)
15.75: yes (hits 16.5)  
21   : no  
26.25: no  
31.5 : no  
36.75: no  
42   : no  
47.25: no  
52.5 : no
57.75: no

So you can see that he hits 3 out of every 11 cracks, or 27.27...% of all cracks he passes. Important to note is that he just barely misses the final crack, at 57.75. But since we defined the problem as him being just past the crack to begin with, we are not counting that first crack (which is actually the last crack of the previous cycle).


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