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Maths/ Logics problems! Options
Etcheparre
Posted: Thursday, February 24, 2011 9:18:28 AM

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Bored at work so here's a fun math problem:
a=1
b=1

so

a=b
a^2=ab
a^2-b^2=ab-b^2
(a-b)(a+b)=(a-b)b
a+b=b
1+1=1
2=1

Where did it go wrong?!

Anyone else got some logics/maths problems which would help kill time?
bluecloud
Posted: Thursday, February 24, 2011 9:36:46 AM

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I don`t know how to explain it in Math terms.

(a-b)(a+b)= (a-b)b

you cannot divide the equation by (a-b) as long as it is equal to 0.

nice topic
pedro
Posted: Thursday, February 24, 2011 10:05:16 AM

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Find five distinct whole numbers between 1 and 10 so that the cubes of three of them add up to the same number as the cubes of the other two.
KenMac
Posted: Thursday, February 24, 2011 10:42:31 AM
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I think it goes wrong at step a+b=b. a+b=2
Etcheparre
Posted: Thursday, February 24, 2011 10:55:33 AM

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bluecloud wrote:
I don`t know how to explain it in Math terms.

(a-b)(a+b)= (a-b)b

you cannot divide the equation by (a-b) as long as it is equal to 0.


Yup that's it, you can't divide by 0 :p

Pedro, you want a solution to a^3+b^3+c^3=d^3+e^3 right, with all the numbers between 1 and 10, and whole?
pedro
Posted: Thursday, February 24, 2011 11:01:33 AM

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Etcheparre wrote:
[quote=bluecloud]

Pedro, you want a solution to a^3+b^3+c^3=d^3+e^3 right, with all the numbers between 1 and 10, and whole?



That's the task
Etcheparre
Posted: Thursday, February 24, 2011 11:11:31 AM

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Tough stuff... God this is going to keep me "working" for a while!
bluecloud
Posted: Thursday, February 24, 2011 11:17:38 AM

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Etcheparre wrote:
Tough stuff... God this is going to keep me "working" for a while!

you`re funny Applause
Etcheparre
Posted: Thursday, February 24, 2011 11:50:17 AM

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pedro wrote:

That's the task


Don't think its possible but I can't prove it... Brick wall
I guess I need to work on congruences and divisibility but that dates way to far back for me to remember... Brick wall Brick wall Brick wall
bluecloud
Posted: Thursday, February 24, 2011 12:00:02 PM

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pedro, give us a clue

or we should ask prof. Stewart:
http://www2.warwick.ac.uk/alumni/knowledge/projects/live/archive/ianstewart/
Winnie
Posted: Thursday, February 24, 2011 12:10:30 PM

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Etcheparre wrote:
pedro wrote:

That's the task


Don't think its possible but I can't prove it... Brick wall
I guess I need to work on congruences and divisibility but that dates way to far back for me to remember... Brick wall Brick wall Brick wall


I think you're overcomplicating things.

What happened to the good old trial and error?

I've already found the answer but if there is another more "logical" way to solve this, I would really like to see it.
bluecloud
Posted: Thursday, February 24, 2011 12:12:46 PM

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tell us the answer, please

I`ve been trying for a long time `the good old trial and error` to no avail
abcxyz
Posted: Thursday, February 24, 2011 12:17:15 PM

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EDITED:

a^3+b^3+c^3=d^3+e^3

Without any loss of generality,
a=3k

b=3p+r, d=3q+r, e=3s+u, c=3t+u

d^3+e^3-(b^3+c^3)=9r^2(q-p)+9u^2(s-t)

p,q,s,t<=3, r,u<3

possible cases are: r=2, s-t=2, u=1=q-p; r=u=1=q-p=t-s; q-p=+-3, u=0;+-(q-p)=r=1, +-(s-t)=u=2

we can show that except the first case that gives the set 8,7,9,1,5, the other 2 cases are impossible.
abcxyz
Posted: Thursday, February 24, 2011 12:22:24 PM

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Winnie wrote:
Etcheparre wrote:
pedro wrote:

That's the task


Don't think its possible but I can't prove it... Brick wall
I guess I need to work on congruences and divisibility but that dates way to far back for me to remember... Brick wall Brick wall Brick wall


I think you're overcomplicating things.

What happened to the good old trial and error?

I've already found the answer but if there is another more "logical" way to solve this, I would really like to see it.


What's the answer? Brick wall
abcxyz
Posted: Thursday, February 24, 2011 12:26:20 PM

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Oh hang on, I made a mistake. The set is 8,7,9,5,1.
Winnie
Posted: Thursday, February 24, 2011 12:27:26 PM

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bluecloud wrote:
tell us the answer, please

I`ve been trying for a long time `the good old trial and error` to no avail


1 5 9 7 8
bluecloud
Posted: Thursday, February 24, 2011 12:31:57 PM

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thank you

next problem, please
abcxyz
Posted: Thursday, February 24, 2011 12:49:53 PM

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bluecloud wrote:
thank you

next problem, please


sorry, a typo. new problem: EDIT: can you draw 11 lines through 9 points so that there are 4 points on each line?
KenMac
Posted: Thursday, February 24, 2011 3:54:34 PM
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a=1
b=1

so

a=b 1=1 (true)
a^2=ab 1 squared = 1 x 1 (still 1=1 true)
a^2-b^2=ab-b^2 1 squared - 1 squared = 1 x 1 - 1 squared (0=0 also true)
(a-b)(a+b)=(a-b)b 0 x 2 = 0 x 1 (still 0=0 true)
a+b=b 2 = 1 (false, or mathematician, please explain)
1+1=1 2 = 1
2=1 2 = 1

I'd like to be daring and bet my 9 points on this.
guitar53
Posted: Thursday, February 24, 2011 4:09:50 PM

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Ms. Green has 40 apples.
She gives 18 to Tommy Jones @ 47... hey! where's the cents symbol?
Jyrkkä Jätkä
Posted: Thursday, February 24, 2011 4:24:17 PM

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guitar53 wrote:
Ms. Green has 40 apples.
She gives 18 to Tommy Jones @ 47... hey! where's the cents symbol?


If you have Windows, alt-0162 = ¢
if Linux, Compose+|+C

guitar53
Posted: Thursday, February 24, 2011 5:26:07 PM

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"Introduction to the new math" left me in the dust when it was introduced in my middle school. I miss Ms. Greens apples and the two trains.
rob
Posted: Thursday, February 24, 2011 8:20:45 PM

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Location: Inside the Blackhole.
To whom it may concern,

I would like to discuss the problem of division by zero in the set of real numbers. So far the best explanation I've found to see why a quotient like a/0 cannot be defined in the set of real numbers is in an old textbook which shows that division by zero is undefined because division is defined by multipication which becomes an identity in the case of the denominator equaling zero.

a/b = c is defined by a = b*c.

"If a/0 = c, then a = 0*c.
But 0*c = 0.
Hence, if a is not equal to 0,
no value of c can make the statment a = 0*c true,
while if a = 0,
every value of c will make the statement true.

Thus, a/0 either has no value or is indefinite in value."


Yours respectfully,
Etcheparre
Posted: Friday, February 25, 2011 5:16:13 AM

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KenMac wrote:
a=1
b=1

so

a=b 1=1 (true)
a^2=ab 1 squared = 1 x 1 (still 1=1 true)
a^2-b^2=ab-b^2 1 squared - 1 squared = 1 x 1 - 1 squared (0=0 also true)
(a-b)(a+b)=(a-b)b 0 x 2 = 0 x 1 (still 0=0 true)
a+b=b 2 = 1 (false, or mathematician, please explain)
1+1=1 2 = 1
2=1 2 = 1

I'd like to be daring and bet my 9 points on this.


Aye! You just can't divide by zero :p

I have another one but I actually have a bit of work today, so I'll post it later on!
Etcheparre
Posted: Friday, February 25, 2011 6:02:58 AM

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Two simple ones:

1. Calculate, without a calculator of course (and for those who know it, might as well avoid Gauss's formula):
1/500+2/500+3/500+4/500+...+498/500+499/500


2.
A = 2000

B = A - 999

C = A + B - 998

D = A + B + C - 997

.

.

.

.

.

Z = A + B + C + .......... + Y - 975

Find Z!
abcxyz
Posted: Friday, February 25, 2011 7:13:13 AM

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1. s=1/500+2/500+3/500+4/500+...+498/500+499/500= (500-1)/500 + ... + (500 - 499)= 499 - 1/500 - 2/500 -..- 499/500
2s = 499
s= 499/2

2. a=2000
b=a-999
c=(a-999)+(a-998)
d=(a-999)+(a-998)+(a-997)
z=(a-999)+...+(a-975)=1001+1002+..+1025=25000+25*13=25325
Etcheparre
Posted: Friday, February 25, 2011 8:49:32 AM

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Yes and nope :)

For the first problem the way I do it is:

1/500+499/500=1
2/500+498/500=1
3/500+497/500=1
.
.
.
248/500+252/500=1
249/500+251/500=1
250/500=1/2

So s=249+1/2=499/2

The way with Gauss is simply 1/500th of the sum of the 499 first whole numbers, which is equal to: 1/500*[(499(500)/2)]=499/2

For the second one look at the way you express d;

This is how it should be expressed:

d=a+b+c-997
d=(a-997)+(a-999)+c
d=(a-997)+(a-999)+[b+(a-998)]
d=(a-997)+(a-999)+[(a-999)+(a-998)]
d=(a-997)+(a-999)+(a-999)+(a-998)
d=2*(a-999)+(a-998)+(a-997)
AND NOT
d=(a-999)+(a-998)+(a-997)

abcxyz
Posted: Friday, February 25, 2011 9:56:55 AM

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Oh shyt.. ok, here goes

z-y=y+1, this holds for all c,..,z.
z=2y+1=2(2x+1)+1=..=2^24b+1+2+..+2^23=2^24(b+1)-1=1024*1024*16*1002-1
Is this correct now? I have a hunch I've made a mistake somewhere again.. oh well
Etcheparre
Posted: Friday, February 25, 2011 11:11:02 AM

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abcxyz wrote:
Oh shyt.. ok, here goes

z-y=y+1, this holds for all c,..,z.
z=2y+1=2(2x+1)+1=..=2^24b+1+2+..+2^23=2^24(b+1)-1=1024*1024*16*1002-1
Is this correct now? I have a hunch I've made a mistake somewhere again.. oh well


I don't actually have the answer but I've been following pretty much the same logic as you have, but haven't found an answer yet...
Etcheparre
Posted: Friday, February 25, 2011 11:19:10 AM

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I'll probably get back to this after the week end :p
Etcheparre
Posted: Friday, February 25, 2011 12:28:13 PM

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abcxyz wrote:
Oh shyt.. ok, here goes

z-y=y+1, this holds for all c,..,z.
z=2y+1=2(2x+1)+1=..=2^24b+1+2+..+2^23=2^24(b+1)-1=1024*1024*16*1002-1
Is this correct now? I have a hunch I've made a mistake somewhere again.. oh well


That's not exactly what I got:

k(n)=2^n*k(0)-1
with a=k(0), so z=k(25)=2^25*k(0)-1=1024*1024*32-1

I'll look into it Monday; have a god weekend ya'll!
abcxyz
Posted: Friday, February 25, 2011 1:02:05 PM

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Etcheparre wrote:
abcxyz wrote:
Oh shyt.. ok, here goes

z-y=y+1, this holds for all c,..,z.
z=2y+1=2(2x+1)+1=..=2^24b+1+2+..+2^23=2^24(b+1)-1=1024*1024*16*1002-1
Is this correct now? I have a hunch I've made a mistake somewhere again.. oh well


That's not exactly what I got:

k(n)=2^n*k(0)-1
with a=k(0), so z=k(25)=2^25*k(0)-1=1024*1024*32-1

I'll look into it Monday; have a god weekend ya'll!


k(n)=2^n*k(0)-1 with a=k(0) does not satisfy when n=1. According to this formula, b=2000*2-1=3999, but a-999=1001.
According to my method it's k(n)=2^(n-1)k(1)+2^(n-2)-1, where k(1)=b.

Have a good weekend :)
abcxyz
Posted: Friday, February 25, 2011 8:52:13 PM

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abcxyz wrote:
Etcheparre wrote:
abcxyz wrote:
Oh shyt.. ok, here goes

z-y=y+1, this holds for all c,..,z.
z=2y+1=2(2x+1)+1=..=2^24b+1+2+..+2^23=2^24(b+1)-1=1024*1024*16*1002-1
Is this correct now? I have a hunch I've made a mistake somewhere again.. oh well


That's not exactly what I got:

k(n)=2^n*k(0)-1
with a=k(0), so z=k(25)=2^25*k(0)-1=1024*1024*32-1

I'll look into it Monday; have a god weekend ya'll!


k(n)=2^n*k(0)-1 with a=k(0) does not satisfy when n=1. According to this formula, b=2000*2-1=3999, but a-999=1001.
According to my method it's k(n)=2^(n-1)k(1)+2^(n-1)-1, where k(1)=b.

Have a good weekend :)
Etcheparre
Posted: Thursday, March 3, 2011 6:42:38 AM

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abcxyz wrote:


k(n)=2^n*k(0)-1 with a=k(0) does not satisfy when n=1. According to this formula, b=2000*2-1=3999, but a-999=1001.
According to my method it's k(n)=2^(n-1)k(1)+2^(n-2)-1, where k(1)=b.

Have a good weekend :)


Wow, what a weekend :p

Yeah my bad, I went running after sums for some reason; your method is waayyy better...
Etcheparre
Posted: Friday, March 11, 2011 6:37:23 AM

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This isn't really a problem, more of a magic trick. Here's how the trick works:
The "magician" asks someone to write a random four digit number on a sheet of paper, anything between 0001 and 9999. Then the "magician" writes a number on another sheet of paper and hides it; he then asks the same person to write another four digit number under the other number. The magician then writes a four digit number under the two other numbers, the "victim" writes another four digit number and the magician then writes another number. The sum of all these numbers is equal to the number he wrote on the other sheet of paper.

For instance:
7898 the magician writes 27898 on the other sheet, shows it to no one
2546
6556 written by the magician
4789
6109 written by the magician
Sum of all five is equal to 27898

I understand how he does it mathematically, what surprises me is how quickly he does it. I've devised a means to make it work but it involves some basic calculus, while this trick was shown to me by a drunken bartender, which most probably didn't do much calculus...

Anyone have a smart method for this?
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