You might have heard of the following problem.
As I was going to St. Ives, I met a man with seven wives. Every wife had seven sacks, every sack had seven cats, every cat had seven kitts. Kitts, cats, sacks, wives, how many were going to St. Ives?
This is a rather good problem and a rather old problem. In fact some of the hieroglyphics found on papyrus more than 3000 years old are believed to represent a very similar problem!
I think this problem, while amusing, is a little big ambiguous and slightly alarming. Is the toothed, clawed, furry and wild-haired army en route to St.Ives like the narrator or is the narrator passing them while they are in their their home? Can we
assume the man meets the army head on as they are returning from St.Ives? Is the man travelling with his entourage or is just meeting them on his way to St.Ives? What kind of person carries 7 cats and 49 kittens in a sack? What...
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Say we were tasked to find the sum of the series from 51 through 375 of series incremented by six. That's a difference of 324 so it would take 324 hops if the hops were each one in size, but the hops are six times bigger than that so it takes six times
fewer hops. 324/6 = (300+24)/6 = 50 + 4 = 54.
So note that there are 55 numbers in the series! Only one of the initial and last numbers corresponds to any of those hops (take your pick, it all depends on your perspective which one). In any case, there is one more stop than hop, considering that
you are already at a stop before the first hop and you'll arrive at a stop after the last hop.
Option A:
We could subtract the first number from every number in the series, and end up with 55 of those subtracted terms.
So we could compute our sum as 55*51 + 6,12,18...318, 324
The first part is as easy as (here's an easy way to multiply any number by 51, write 51 as...
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