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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^2b^2=abb^2 (ab)(a+b)=(ab)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?


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I don`t know how to explain it in Math terms.
(ab)(a+b)= (ab)b
you cannot divide the equation by (ab) as long as it is equal to 0.
nice topic


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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.


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I think it goes wrong at step a+b=b. a+b=2


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bluecloud wrote:I don`t know how to explain it in Math terms.
(ab)(a+b)= (ab)b
you cannot divide the equation by (ab) 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?


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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


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


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


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pedro wrote: That's the task
Don't think its possible but I can't prove it... I guess I need to work on congruences and divisibility but that dates way to far back for me to remember...


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Etcheparre wrote:pedro wrote: That's the task
Don't think its possible but I can't prove it... I guess I need to work on congruences and divisibility but that dates way to far back for me to remember... 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.


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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


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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(qp)+9u^2(st)
p,q,s,t<=3, r,u<3
possible cases are: r=2, st=2, u=1=qp; r=u=1=qp=ts; qp=+3, u=0;+(qp)=r=1, +(st)=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.


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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... I guess I need to work on congruences and divisibility but that dates way to far back for me to remember... 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?


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Oh hang on, I made a mistake. The set is 8,7,9,5,1.


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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


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thank you
next problem, please


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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?


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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^2b^2=abb^2 1 squared  1 squared = 1 x 1  1 squared (0=0 also true) (ab)(a+b)=(ab)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.


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


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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, alt0162 = ¢ if Linux, Compose++C


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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.


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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,


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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^2b^2=abb^2 1 squared  1 squared = 1 x 1  1 squared (0=0 also true) (ab)(a+b)=(ab)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!


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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!


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1. s=1/500+2/500+3/500+4/500+...+498/500+499/500= (5001)/500 + ... + (500  499)= 499  1/500  2/500 .. 499/500 2s = 499 s= 499/2
2. a=2000 b=a999 c=(a999)+(a998) d=(a999)+(a998)+(a997) z=(a999)+...+(a975)=1001+1002+..+1025=25000+25*13=25325


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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+c997 d=(a997)+(a999)+c d=(a997)+(a999)+[b+(a998)] d=(a997)+(a999)+[(a999)+(a998)] d=(a997)+(a999)+(a999)+(a998) d=2*(a999)+(a998)+(a997) AND NOT d=(a999)+(a998)+(a997)


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Oh shyt.. ok, here goes
zy=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*10021 Is this correct now? I have a hunch I've made a mistake somewhere again.. oh well


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abcxyz wrote:Oh shyt.. ok, here goes
zy=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*10021 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...


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I'll probably get back to this after the week end :p


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abcxyz wrote:Oh shyt.. ok, here goes
zy=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*10021 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*321 I'll look into it Monday; have a god weekend ya'll!


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Etcheparre wrote:abcxyz wrote:Oh shyt.. ok, here goes
zy=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*10021 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*321 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*21=3999, but a999=1001. According to my method it's k(n)=2^(n1)k(1)+2^(n2)1, where k(1)=b. Have a good weekend :)


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abcxyz wrote:Etcheparre wrote:abcxyz wrote:Oh shyt.. ok, here goes
zy=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*10021 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*321 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*21=3999, but a999=1001. According to my method it's k(n)=2^(n1)k(1)+2^(n1)1, where k(1)=b. Have a good weekend :)


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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*21=3999, but a999=1001. According to my method it's k(n)=2^(n1)k(1)+2^(n2)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...


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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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