Zeno's Other Frog

1+2+ . . . +2^(n-1)>=10^100 =>2^n-1=10^100 100/log(2) (base 10) is little over 332 but less than 333 so answer is 333

I did this with help from excel, made 3 columns. 1 to track the # of jumps, 1 to track the distance of the current jump and 1 to track the total distance. I noticed a pattern pretty quickly as well.

Jump 2 was a distance of 2ft and a total distance of 3ft. Jump 3 was a distance of 4ft and a total distance of 7ft. Jump 4 was a distance of 8ft and a total distance of 15ft. Jump 5 was a distance of 16ft and a total distance of 31ft. Jump 6 was a distance of 32ft and a total distance of 63ft.

The total distance is always 1ft less than the length of the current jump x2. After noticing that I was able to drag the formula down until the jump distance was over half a googol, which happens at jump 333.

Approached with 10 lines of Python:

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googol = 10**100
traveled = 0
jump_distance = 1

for i in xrange(1,1500):
    traveled += jump_distance
    jump_distance *= 2
    if traveled >= googol:
        print "The frog is beyond googol."
        print 'hop:',i
        break

2^jumps = 10^100

333 to exceed a googel