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Fun math!

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Fun math!

Postby Riss » Wed Oct 08, 2008 6:39 pm

A few math problems for the Asylum...
Quite the challenge, I see people plowing through all sorts of math with Wrath just around the corner, and a bunch of theory-crafting. So I thought, why not share some math fun ^^

First problem.
Figure out what division problem was worked:
(Each X represents a digit – not necessarily the same digit – none of the numbers “start” with 0)

Image


Second Problem.
Each of the following letters stands for a different digit (no zeros were used). Find the two possible values for E:
Image
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Postby Dorvan » Wed Oct 08, 2008 6:42 pm

The first problem doesn't make sense as written, are there supposed to be other X's surrounding the 8?
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Postby Elsie » Wed Oct 08, 2008 6:59 pm

Step 1: none of the numbers start with 0

ie, xxx >= 100, xx,xxx,xxx >= 10,000,000

So, assuming 8 = 800, what 3 digit number divides an 8 digit number 800 times?
(if 8 = 8 we can assume the last two XX are decimal places?)
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Postby Dorvan » Wed Oct 08, 2008 7:06 pm

Well yeah, that's what I'm seeking clarification on: is "8" 8, 800, 8XX, or any number with 8 in the hundreds place?
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Re: Fun math!

Postby Talaii » Wed Oct 08, 2008 7:11 pm

Riss wrote:Second Problem.
Each of the following letters stands for a different digit (no zeros were used). Find the two possible values for E:
Image


E = 6 or E = 8.
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Postby Khisvir » Wed Oct 08, 2008 7:15 pm

Dorvan wrote:any number with 8 in the hundreds place


...would be my guess.

Just from the size of the numbers involved, the solution to the division problem is going to be bigger than 800
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Postby Invisusira » Wed Oct 08, 2008 7:15 pm

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Postby Elsie » Wed Oct 08, 2008 7:26 pm

Dorvan wrote:Well yeah, that's what I'm seeking clarification on: is "8" 8, 800, 8XX, or any number with 8 in the hundreds place?

The weirder part is it goes into it first 3 times.

If we assume 0s infront of 8:
That implies xxx goes into the first four, XXXX 0 times.
ie, xxx > XXXX
Therefore, there has to be X's infront of the 8 since there is a remainder in the picture.

(still completely bunk to give a problem like that with only 8 given, and X where any given integer would be. Mind problems where "Haha you made an assumption" are a failure if the assumption is due to giver's lack of commication skills. ie, it is a fault of the asker, not answerer).

So, if there have to be numbers infront of 8, then we have some knowledge:
It xxx goes into XXXX N times.
xxx*N = XXXX+ab (where a !=0)
This number, with the next two XX stated as cd, give abcd.
xxx goes into abcd 8 times.
-> find all situations where xxx*abcd = 8 remainder ZY

Also we know the number directly before 8 is 0, otherwise we'd be doing abc instead of abcd.
=>ergo, answer is N08 so far.

I've got a feeling that XXXX = abcd since they both give two remainder digits
We Assume: a!=0, xxx >=100, N = 8.
We Know: xxx divides abcdcd
We Know: xxx divides abcd 8 times remainder ab

This actually gives us a lot of information.
Firstly: xxx*8 + ab = abcd
=> xxx*8 = abcd +ab
=> xxx = (abcd + ab) / 8

Sadly, there's 5832 combinations for abcd, and 72 for just ab. we do know that b and d are both even or both odd.
We also know that d+b /8 = the last digit in xxx, which from now on will be xyz. => z = d+b/8
We know d, b < 10, so d+b ranges 0-18, so d+b /8 = 0, 1, or 2 => d+b = 0, 8, or 16
If z is 0, then d,b,f is 0. (f is defined below)
If z is 1, then d,b are..
if z is 2, then d,b, are...


We also know that, under the assumption XXXX = abcd, there is a four digit number abef where xyz divides abef evenly.

This should get someone else started. I have a Management test tomorrow and a paper due in about 9 hours.
I'm working with xyz (divides) abcdcdef
xyz*N = abcd+ab
xyz*Y = abef + 0
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Postby Riss » Wed Oct 08, 2008 8:40 pm

For the record these problems are directly from my teacher, word for word and the images are exactly as shown on the handout. Some crazy college professors...
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Postby Elsie » Wed Oct 08, 2008 9:41 pm

Riss wrote:For the record these problems are directly from my teacher, word for word and the images are exactly as shown on the handout. Some crazy college professors...

It's not a bad question, it's just poor form to state a question where all X's are integers. This leads to the assumption that integers only exist where X's do.

The problem is solvable so long as 8 isn't the first number. It isn't particularly hard - I just forgot the exact method so I would need to go through all the possible information.

I mean, we know the final number -has- to be X080X without a remainder. There's only 72 possibilities there.
There's also only so many ways to get XXXX/xxx = X remainder XX.
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Postby mazater » Wed Oct 08, 2008 11:23 pm

PROOF THAT GIRLS ARE EVIL

Girls = Time * Money <given>

Time = Money <given>

Girls = Money * Money <substitution>

Girls = Money^2 <algebra>

Money = Evil^(1/2) <given>

therefore: Girls = [Evil^(1/2)]^2 <substitution>

Girls = Evil Q.E.D. <algebra>

Maths, oh how I love thee.
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Postby anthony » Thu Oct 09, 2008 9:17 am

Mazater wrote:PROOF THAT GIRLS ARE EVIL

Girls = Time * Money <given>

Time = Money <given>

Girls = Money * Money <substitution>

Girls = Money^2 <algebra>

Money = Evil^(1/2) <given>

therefore: Girls = [Evil^(1/2)]^2 <substitution>

Girls = Evil Q.E.D. <algebra>

Maths, oh how I love thee.


yet maths shall not stop men from wanting them.
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Postby Omatre » Thu Oct 09, 2008 10:15 am

As the story goes:

The part of a woman a man wants is the vagina.
The rest of it is the cunt that its wrapped up in that you gotta suffer for.
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Postby duruk » Fri Oct 10, 2008 12:28 am

Invisusira wrote:Image


I got this printed out on my desk. All the professors and Ph.D. giggle at it. They also think it's great for it's simplicity, and it reminds them of keeping it simple :)

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Postby Andurin » Fri Oct 10, 2008 7:20 am

Try this one aswell then:

Image
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