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December 30th, 2018, 10:29 AM   #1
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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 + n-squared + n-cubed + 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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December 30th, 2018, 11:15 AM   #2
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Quote:
Originally Posted by RichardJ View Post
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.
Simple "luck": 7 + 49 + 343 + 2401 = 2800

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!!

Last edited by Denis; December 30th, 2018 at 11:25 AM.
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December 30th, 2018, 11:22 AM   #3
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Originally Posted by Denis View Post
Simple "luck": 7 + 49 + 343 + 2401 = 2800
$\sum \limits_{k=1}^4~7^k = \dfrac{1-7^{5}}{1-7} -1 = 2800$
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December 30th, 2018, 12:20 PM   #4
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December 30th, 2018, 12:42 PM   #5
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ahhh so it's actually 2801

2800.... and then came Bronson
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December 30th, 2018, 01:09 PM   #6
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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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December 30th, 2018, 02:14 PM   #7
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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.
Hokay, Hokay; so you don't lose any sleep, let's clarify:

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.

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December 30th, 2018, 02:19 PM   #8
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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^5-x)/(x-1). 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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December 30th, 2018, 03:08 PM   #9
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Quote:
Originally Posted by Denis View Post
Hokay, Hokay; so you don't lose any sleep, let's clarify:

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.

I dunno. If you leave St. Ives going due east you will eventually arrive back at St. Ives. So I don't know that your solution works. Just sayin'.
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December 30th, 2018, 04:55 PM   #10
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Quote:
Originally Posted by Denis View Post
Hokay, Hokay; so you don't lose any sleep, let's clarify:

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.

He was going east since he forgot his cellphone at the last resting phone. Once he found it again, he'll resume to St. Ives
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