Author Topic: The Jesus Christ Problem  (Read 6396 times)

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Offline Ender

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The Jesus Christ Problem
« on: November 23, 2006, 07:59:19 pm »
The customer favorite in the Joe's Nicotine-Enhanced Candy Store is the Jesus Christ ice cream bar. Jesus Christ Ice Cream Bars come in either packs of 7 or packs of 9.

Q1. What is the maximum number of Jesus Christ Ice Cream Bars that one cannot buy?
Q2. How many integer amounts of Jesus Christ Ice Cream Bars can one not buy?
Q3. Generalize your answer to Q1 to any situation where Jesus Christ Ice Cream bars come in either packs of a or packs of b, where a and b are integers.

You don't have to wait until you get solutions to all three to post.

Offline rabbit

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Re: The Jesus Christ Problem
« Reply #1 on: November 23, 2006, 09:18:30 pm »
1. infinite
2. infinite
3. yes

Offline Ender

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Re: The Jesus Christ Problem
« Reply #2 on: November 23, 2006, 10:56:01 pm »
Incorrect.

Offline Nate

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Re: The Jesus Christ Problem
« Reply #3 on: November 25, 2006, 10:40:25 pm »
You worded something wrong because the first two are infinite.

Offline Joe

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Re: The Jesus Christ Problem
« Reply #4 on: November 25, 2006, 11:44:08 pm »
You worded something wrong because the first two are infinite.

Or he just said that at least one of them was incorrect, that being 3. It doesn't even make sense.
I'd personally do as Joe suggests

You might be right about that, Joe.


Offline leet_muffin

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Re: The Jesus Christ Problem
« Reply #5 on: November 26, 2006, 06:43:27 pm »
1. 47?
The douchebag method:
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Offline d&q

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Re: The Jesus Christ Problem
« Reply #6 on: November 26, 2006, 07:29:33 pm »
The first two are infinite, one cannot buy an amount who's prime factorization does not include 9 or 7, which is an infinite set of numbers. I'm not sure about the third one though. Leet_muffin, how did you get 47?
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Offline leet_muffin

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Re: The Jesus Christ Problem
« Reply #7 on: November 26, 2006, 08:04:45 pm »
Try and add combinations of 9 and 7 to get 47. I couldn't. I didn't care to go any higher.
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Offline d&q

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Re: The Jesus Christ Problem
« Reply #8 on: November 26, 2006, 09:01:11 pm »
47 is a prime number, you shouldn't have bothered.  :P

[Edit]: Also, there is an infinitely high number of prime numbers if you care to know. I could look up the proof but I'm eating chicken right now and I'm too lazy.
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Offline leet_muffin

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Re: The Jesus Christ Problem
« Reply #9 on: November 26, 2006, 09:43:16 pm »
This has nothing to do with prime numbers...
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Offline d&q

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Re: The Jesus Christ Problem
« Reply #10 on: November 26, 2006, 11:42:58 pm »
How does it not?

Try and add combinations of 9 and 7 to get 47.

Ex: 9 + 9 +9 + 7 + 7

That is equivalent to 3^3 + 7^2. Any combinations you are adding will be equivalent to some form 3^x + 7^x. Which is called prime factorization.

[Edit]: I misread it, my mistake.  :-\
« Last Edit: November 27, 2006, 12:29:54 am by Deuce »
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Offline leet_muffin

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Re: The Jesus Christ Problem
« Reply #11 on: November 27, 2006, 12:24:08 am »
23 is a prime number. 7+7+9 = 23. I'm still not seeing how it involves prime numbers at all. Can you please explain?
The douchebag method:
fuck allfo you i dont give a fuck ill fight everyone of you fuck that sbhit fuck you

Offline Ender

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Re: The Jesus Christ Problem
« Reply #12 on: December 01, 2006, 06:09:27 pm »
1. 47?

You're right. Explain your answer though :P

Questions 2 and 3 still remain.

Offline leet_muffin

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Re: The Jesus Christ Problem
« Reply #13 on: December 02, 2006, 04:22:15 pm »
Uh, any number (edit:)higher than 47 can be created through different combinations of 9 and 7.

I'll work on 2 and 3 later, I be going afk.
« Last Edit: December 03, 2006, 02:45:21 pm by leet_muffin »
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Offline Ender

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Re: The Jesus Christ Problem
« Reply #14 on: December 03, 2006, 12:58:30 pm »
Uh, any number of 47 can be created through different combinations of 9 and 7.

I'll work on 2 and 3 later, I be going afk.

You need to elucidate. 47 is the maximum number that cannot be created by any combination of 9 and 7.

Offline Miamiandy

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Re: The Jesus Christ Problem
« Reply #15 on: December 07, 2006, 09:35:55 pm »
I got all the answers.

1. The maximum number of bars that cannot be obtained is 47.
2. The number of interger amounts of bars that one cannot buy is 24.
3. If a and b are coprime positive integers then the maximum number of bars that cannot be bought is (a*b-a-b) bars.  Otherwise, an arbitrarily large number of bars cannot be bought.
« Last Edit: December 08, 2006, 07:17:21 am by Miamiandy »
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