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Old 2016-03-14, 16:35   #1
richs
 
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Default Pizza Hut Meth Contest

Win pizza for 3.14 years:

1. I’m thinking of a ten-digit integer whose digits are all distinct. It happens that the number formed by the first n of them is divisible by n for each n from 1 to 10. What is my number?

2. Our school’s puzzle-club meets in one of the schoolrooms every Friday after school.

Last Friday, one of the members said, “I’ve hidden a list of numbers in this envelope that add up to the number of this room.” A girl said, “That’s obviously not enough information to determine the number of the room. If you told us the number of numbers in the envelope and their product, would that be enough to work them all out?”

He (after scribbling for some time): “No.” She (after scribbling for some more time): “well, at least I’ve worked out their product.”

What is the number of the school room we meet in?”

3. My key-rings are metal circles of diameter about two inches. They are all linked together in a strange jumble, so that try as I might, I can’t tell any pair from any other pair.

However, I can tell some triple from other triples, even though I’ve never been able to distinguish left from right. What are the possible numbers of key-rings in this jumble?
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Old 2016-03-14, 16:41   #2
Dubslow
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1 is trivial to brute force.

2 reminds me of the similar problem about the Russian mathematician's sons' age. Presumably it's a similar sort of answer.

I don't even understand 3.
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Old 2016-03-15, 04:36   #3
a1call
 
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Quote:
Originally Posted by richs View Post
Win pizza for 3.14 years:

1. I’m thinking of a ten-digit integer whose digits are all distinct. It happens that the number formed by the first n of them is divisible by n for each n from 1 to 10. What is my number?
Here is 1 answer, I am sure there are others:

9216543870

Last fiddled with by a1call on 2016-03-15 at 04:39
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Old 2016-03-15, 07:50   #4
axn
 
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Quote:
Originally Posted by a1call View Post
Here is 1 answer, I am sure there are others:

9216543870
92165438 is not divisible by 8

There is only one solution, and is of the format OEOE5EOE0, which means you have to brute force 4!*4! = 576 permutations.
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Old 2016-03-15, 09:15   #5
0PolarBearsHere
 
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Quote:
Originally Posted by axn View Post
92165438 is not divisible by 8

There is only one solution, and is of the format OEOE5EOE0, which means you have to brute force 4!*4! = 576 permutations.
Well...
OEOE5EOEO0
But yes, 4!*4!
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Old 2016-03-15, 09:41   #6
axn
 
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Quote:
Originally Posted by 0PolarBearsHere View Post
Well...
OEOE5EOEO0
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Old 2016-03-15, 11:10   #7
a1call
 
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Quote:
Originally Posted by axn View Post
92165438 is not divisible by 8

There is only one solution, and is of the format OEOE5EOE0, which means you have to brute force 4!*4! = 576 permutations.
Excel roundung and late night calculations again,

How about

3816547290

I verified with iPhone calculator.
Thanks for the reply and checking,
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Old 2016-03-15, 16:11   #8
WMHalsdorf
 
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For the first problem a for a brute force you would need to check if the number is a multiples of 2520 (the least common multipler of 1 thru 9).
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Old 2016-03-15, 17:41   #9
a1call
 
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Quote:
Originally Posted by WMHalsdorf View Post
For the first problem a for a brute force you would need to check if the number is a multiples of 2520 (the least common multipler of 1 thru 9).
3816547290 is not a multiple of 2520, yet it satisfies the constraints of the puzzle.
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Old 2016-03-15, 18:59   #10
WMHalsdorf
 
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Quote:
Originally Posted by a1call View Post
3816547290 is not a multiple of 2520, yet it satisfies the constraints of the puzzle.
3816547290 is not divisible by 7

1234759680 meets all constraints. Note there may be larger numbers.
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Old 2016-03-15, 19:02   #11
science_man_88
 
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Quote:
Originally Posted by WMHalsdorf View Post
3816547290 is not divisible by 7

1234759680 meets all constraints. Note there may be larger numbers.

for the first thing only the first 7 digits need to be admittedly they aren't either but it's a key distinction. no your answer doesn't a number ending in 7 does not divide by 5. look at the new post a1call made it seems i was confusing your new number and their old one.

Last fiddled with by science_man_88 on 2016-03-15 at 19:24
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