St. Ives Was thinking about the old rhyme "as I was going to St. Ives," and noticed something odd. For those not familiar the rhyme is: As I was going to St. Ives I met a man with 7 wives Each wife had 7 sacks Each sack had 7 cats Each cat had 7 kits. Kits, cats, sacks and wives, how many were going to St. Ives? The trick answer is 'one,' since presumably only the narrator was going to St. Ives and the others were headed the other way. But... If you do the actual calculation on kits, cats, etc., the answer is 2800. 4 x 7 x 100. Hmmm. That even a number just struck me as oddly coincidental. There has to be some pattern behind that, but if there is, it's completely eluding me. But there's more. I wondered  what it would be if there were 3 wives with 3 sacks each, etc.. The answer to that one is 120. And if it had been 5 or 9, both of those totals also end in 0. In fact for any odd number, other than those ending in 1, that sum of n + nsquared + ncubed + n(to the 4th) will always end in zero. There has to be a reason behind that pattern, but I'm just not seeing it. Anyone feel like tackling that with some kind of general explanation or analysis? 
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Looks like that in order to find more of these: a = 1st term n = number of terms a^1 + a^2 + a^3........+ a^n = u*100 In other words, a complete waste of time:) 2nd edit: thanks Romsek!! 
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ahhh so it's actually 2801 2800.... and then came Bronson 
The "standard" solution of the riddle is dumb. Nowhere is it said that they are going the other way, it just says you met them. Why does that mean they were going the other way. Can't it be that they were taking a break and you walked passed them for example. Or they were going slow and you caught up to them. 
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As I was going West to St. Ives I met a man going East with 7 wives Each wife had 7 sacks Each sack had 7 cats Each cat had 7 kits. Kits, cats, sacks and wives, how many were going to St. Ives? AND answer from now on is NONE, since they're all going East. :p 
It's closely related to the fact that for every integer x, the last digit of x^5 is the same as the last digit of x. (This is a consequence of Fermat's Little Theorem, if you're familiar with that.) In more detail: Being divisible by 10 means divisible by 2 and divisible by 5. It's not too hard to see that x + x^2 + x^3 + x^4 is always divisible by 2. Divisibility by 5 is a bit trickier, but you can use the formula for sums of terms in geometric progression to get x + x^2 + x^3 + x^4 = (x^5x)/(x1). The numerator is always divisible by 5 by the fact mentioned at the start of this post. So we're done, provided that the denominator is not also divisible by 5 (which could cancel with the 5 in the numerator). So this works as long as x  1 is not a multiple of 5, i.e. it works for any natural number x not ending in 1 or 6 (which is slightly more general than the pattern you noticed). If you're willing to get a bit more crazy, you can use similar reasoning to show such things as: For every natural number x ending in 3, 7, or 9, the sum x + x^2 + x^3 + ... + x^20 will be divisible by 100. 
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