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03-30-2011, 10:27 PM
#676
Registered User
Yeah, good one.
s=9
e=5
n=6
d=7
m=1
o=0
r=8
y=2
_9567
+1085
10653
I'm guessing there aren't any other solutions, but I didn't actually try it out for r=(something besides 8).
Last edited by billl; 03-30-2011 at 10:30 PM.
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03-30-2011, 10:53 PM
#677
Registered User
OK, unless I was dumb enough to triple-check my solution to the previous one incorrectly three straight times, here's the next puzzle, which it is my responsibility to provide.
The Wogged Pursuit of Perfection
In a completely different bubble of the Multi-verse that we lovingly refer to as "M", there is another universe with a similar type of cosmology, ruled over (created, in fact) by a deity known as "Gow". (But he's always been "Pat" to his deity-type-friends.)
Anyhow, Pat was never much for surprises, variety, or anything like that, but he was VERY good at one thing, and that was taking care of business promptly. For Pat, creating an entire planet in a week was a little behind the pace--he managed to create whole planetary systems in that much time. When he finally settled down to business, this is how it went:
For his first solar system, he built one star (sun) with one planet that had one moon. The next week, he built another star (always just the one star per system), but this time he was able to give it two planets, each with one moon. In the next system he built, he put three planets around the star, each planet having its own boring ONE moon orbiting it. And Pat stuck to this formula, oh yes, rigidly, steadily producing solar systems that were gradually (and very predictably) larger than previous ones.
After one year (weeks and years are "Earth years" calculated via atomic clock, for the purposes of this puzzle) Pat stopped. How many planetary bodies (stars, planets, and moons) combined did he have after 52 weeks of building them?
Last edited by billl; 03-31-2011 at 12:19 AM.
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03-31-2011, 02:25 AM
#678
Registered User
At the half way point on week twenty six he will make 1 sun 26 planets and 26 moons and that is his average output for the year. Thats 53 bodies times 52 weeks, which is 2756. However my maths prowess is legendary - For all the wrong reasons - so I've probably done it wrong.
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03-31-2011, 02:35 AM
#679
Registered User
Yes, not quite right. You've shown all the necessary brilliance, and have made a nice presentation of the technique that makes it all so elegant, of course. Just a little bit off. (Lucky for me!)
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03-31-2011, 04:05 AM
#680
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03-31-2011, 10:25 AM
#681
Registered User
Yep. (Add first week and last week, then multiply by 26.) Your turn again, jajdude.
Last edited by billl; 03-31-2011 at 10:28 AM.
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03-31-2011, 10:38 AM
#682
If mick likes, he can find one. I got nothing right now. Can do later with more time to spare.
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04-01-2011, 06:33 AM
#683
OK, a cryptogram:
HSI KZD JST HSCDNE SI NDTJE CH ZPP SZE KIOIPX EHTGGIV HSCDNCDA.
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04-01-2011, 02:31 PM
#684
Registered User

Originally Posted by
billl
Yep. (Add first week and last week, then multiply by 26.) Your turn again, jajdude.
I mean - why does that work? How can you possibly know to do that?
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04-01-2011, 02:37 PM
#685

Originally Posted by
prendrelemick
I mean - why does that work? How can you possibly know to do that?
I don't know whether you're taking the mick, but in case you're not, imagine having to add the numbers from one to five.
* ** *** **** *****
The middle one is the average of the others. Or, to put it another way, you could take two from the last one, and put it on the first one. And one from the fourth one and put it on the second one. You make them all the same because they 'balance' round the middle one.
*** *** *** *** ***
So the sum is the middle number (3) times the number of instances (5).
Same principle with the suns and planets. Figure out how many there are in the middle one and multiply by the number of instances.
Last edited by MarkBastable; 04-01-2011 at 02:43 PM.
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04-01-2011, 02:48 PM
#686
Registered User
Except 52 is an even number of elements (with no 'middle' integer--26 is actually the last member of the first half), so one has to take the other (similar) route (summing the extremes and dividing by two to find the average).
Last edited by billl; 04-01-2011 at 02:52 PM.
Reason: not nearly enough parentheses
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04-01-2011, 02:53 PM
#687
[no title]

Originally Posted by
MarkBastable
I don't know whether you're taking the mick, but in case you're not, imagine having to add the numbers from one to five.
* ** *** **** *****
The middle one is the average of the others. Or, to put it another way, you could take two from the last one, and put it on the first one. And one from the fourth one and put it on the second one. You make them all the same because they 'balance' round the middle one.
*** *** *** *** ***
So the sum is the middle number (3) times the number of instances (5).
Same principle with the suns and planets. Figure out how many there are in the middle one and multiply by the number of instances.
Of course, 26 would be the middle of 51, not 52.
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04-01-2011, 02:54 PM
#688
[no title]
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04-01-2011, 07:54 PM
#689

Originally Posted by
prendrelemick
I mean - why does that work? How can you possibly know to do that?
Once read a story about Gauss, a great mathematician. Teacher wanted to keep kids busy so he/she told them to add numbers 1 to 100. He solved it real fast. I believe he, at maybe 6 years old, took out 100 and the middle number 50. Then it's 99+1 + 98 + 2 + 97 +3 ... or 49 x 100. Or was it 101 x 50, what ever worked. This one had 1 + 2 + ... +52 , or 53 x 26 (twice) as you noted, but forgot to add the stars.
http://www.jimloy.com/algebra/gauss.htm
Last edited by jajdude; 04-01-2011 at 08:04 PM.
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04-01-2011, 09:08 PM
#690
Registered User

Originally Posted by
jajdude
ok, a cryptogram:
Hsi kzd jst hscdne si ndtje ch zpp sze kioipx ehtggiv hscdncda.
zdtdxkr*h?
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