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# Solve this mystery

imagefree
Hello friends!!

Here is a mathematical problem. There very basic error in it but almost 90% persons cant solve it.
So, i am posting it here to see weather you can solve!

Animal
The error is on line 3:
(a+b)(a+b) = c(a+b) becomes a^2 + b^2 + 2ab = ca + cb

In your example, you then remove the 2ab term completely by placing half of it on each side of the equality sign. This is where the error lies.

Using your example terms, line 4 states that a^2 + b^2 = ca + cb (note the absence of any ab term). When you input your number (a=2000, b=4000 and c=6000), this term becomes 4,000,000 + 16,000,000 = 12,000,000 + 24,000,000. Clearly this is where the problem lies.

Using the full expression: a^2 + b^2 + 2ab = ca + cb, this becomes 4,000,000 + 16,000,000 + 16,000,000 = 12,000,000 + 24,000,000 which is correct.
imagefree
 Animal wrote: The error is on line 3: (a+b)(a+b) = c(a+b) becomes a^2 + b^2 + 2ab = ca + cb In your example, you then remove the 2ab term completely by placing half of it on each side of the equality sign. This is where the error lies. Using your example terms, line 4 states that a^2 + b^2 = ca + cb (note the absence of any ab term). When you input your number (a=2000, b=4000 and c=6000), this term becomes 4,000,000 + 16,000,000 = 12,000,000 + 24,000,000. Clearly this is where the problem lies. Using the full expression: a^2 + b^2 + 2ab = ca + cb, this becomes 4,000,000 + 16,000,000 + 16,000,000 = 12,000,000 + 24,000,000 which is correct.

Let me say You are absolutely wrong.

If i am moving half of 2ab to the other sife of Equality sign, then what is the problem? I am not violating any mathematical rule because while taking ab to the other side, i have reversed it sign to negative. So, error is somewhere else.

the equalities
aa+ab+ab+bb=ca+cb and
aa+ab-ac=-ab-bb+bc are perfect alternative of each other. Try putting values in these equalities, you will get same result on both side of equalities.

One more thing is that i have never used the expression a^2 + b^2 + 2ab = ca + cb in the Problem.

So Guys!!! Solution is Still to come from your side. It is Guaranteed that problem Exists
rohan2kool
the problem is simple and obviously in your last step. Here, you cancel (a+b-c) and (a+b-c).. Now, as i say cancel I mean, you consider (a+b-c)/(a+b-c) = 1. But actually, a+b-c evaluates to 2000 + 4000 - 6000. Which evaluates to 0. Now, the fact that q/q = 1 is true for the domain R - {0}.

That is because 0/0 is an indeterminate form. It's value is unknown. For example,

f(x) = (x^x + x)/(x)

then at x = 0, f(x) = 0/0
but as x -> 0, f(x) -> 1

but,

f(x) = (x^x + 2*x)/(x)

then at x = 0, f(x) = 0/0
but as x -> 0, f(x) -> 2

consider:

f(x) = cos(x)/(x - π/2)

at x = π/2, f(x) = 0/0
but as x -> π/2, f(x) -> -1

One can also see this as:

5 x 2 = 5 x 2

therefore 5 = 5 holds true.

but 15 x 0 = 5 x 0

therefore 15 = 5, is not true.

and it's got nothing to do with big numbers playing on your psychology. It would work perfectly on any numbers a, b, c such that (a+b) = c

Refer:

http://en.wikipedia.org/wiki/Indeterminate_form
Captain Fertile
I hate problems like this, it twists my melon and reminds me how dumb I really am.

I tried to work it out but my head exploded!!!!
imagefree
 rohan2kool wrote: the problem is simple and obviously in your last step. Here, you cancel (a+b-c) and (a+b-c).. Now, as i say cancel I mean, you consider (a+b-c)/(a+b-c) = 1. But actually, a+b-c evaluates to 2000 + 4000 - 6000. Which evaluates to 0. Now, the fact that q/q = 1 is true for the domain R - {0}. That is because 0/0 is an indeterminate form. It's value is unknown. For example, f(x) = (x^x + x)/(x) then at x = 0, f(x) = 0/0 but as x -> 0, f(x) -> 1 but, f(x) = (x^x + 2*x)/(x) then at x = 0, f(x) = 0/0 but as x -> 0, f(x) -> 2 consider: f(x) = cos(x)/(x - π/2) at x = π/2, f(x) = 0/0 but as x -> π/2, f(x) -> -1 One can also see this as: 5 x 2 = 5 x 2 therefore 5 = 5 holds true. but 15 x 0 = 5 x 0 therefore 15 = 5, is not true. and it's got nothing to do with big numbers playing on your psychology. It would work perfectly on any numbers a, b, c such that (a+b) = c Refer: http://en.wikipedia.org/wiki/Indeterminate_form

Yes it is. You got actual problem.
You did a lot of research.
rohan2kool
 imagefree wrote: Yes it is. You got actual problem. You did a lot of research.

Nope.. not a lot of research.. just high school algebra.
IceCreamTruck
If he got the answer right then what does he win? a cookie?

My head exploded too, and I wasn't able to find the problem. Not because of the lack of brain power to find where the fault lies in your crazy math problem, but because of a lack of will power to do so considdering no prize was to be awared the winner of the "contest".

I now go elsewhere in search of fame, glory, and power over the masses and a laugh at your attempt to make me spend hours crunching useless numbers.

 Randall from the movie Clerks wrote: I don't appreciate your rouse ma'am.... Your cunning attempt to trick me.
imagefree
rohan2kool wrote:
 imagefree wrote: Yes it is. You got actual problem. You did a lot of research.

Nope.. not a lot of research.. just high school algebra.

blinx
I just did it (same way as the rohan2cool) and I'm 15. I'm pretty goos at maths though, I suppose...