Hi guys!

Here is a math quiz.
I was carrying a stone weighing 40 kgs. As I was walking thru the road( chilpo bato,rat mato) , my feet slipped and I fell down. The stone broke into 4 pieces. To my suprise, it broke into such pieces that now with the combination of those four pieces of stone , I can weigh any integer number of kgs from 1 to 40. To elaborate, i can weigh 1, 2, 3 ... 40 kgs of rice with the combination of those stones.
U guys need to find out the weights of those four stones...Mathematical proof of the answer will be truely welcomed.

Hey Kiwi,
I agree that the question I gave earlier was of bit high standard. Thought people would try. Anyway?. will come up with easier ones next time.


Here is the mathematical proof for your question. It looks ok to me.

Let a, b, c, d be the 4 pieces in which the weight is divided.
where a < b< c< d.

The first equation will be:
a + b+ c+d = 40------ (1) (sum of all 4 pieces).

The second equation will be:
(a +b+ c) = d ? (a +b+ c) -1 (The sum of the first three pieces should be 1 less than the difference between the last piece and the sum of the first three pieces. They can?t be equal other wise we will have one particular weight with two combinations).
Simplifying that further, we get,
2(a +b+ c) = d - 1 -------- (2)

The third equation will be:
(a + b) = c ? (a + b) ? 1 (same reason as above. The sum of the first two pieces should be 1 less than the difference between the third piece and the sum of the first two pieces.
Simplifying,
2 (a + b) = c ? 1------- (3)

The fourth equation will be:
a = b ?a ? 1 (same reason as above)
2a = b-1--------------- (4)

There is a nice pattern in these equations.

Plug in equation (2) in (1), you get d = 27.
Plug in equation (3) in (1), you get c = 9 (By this time, you already know that d = 27).
Plug in equation (4) in (1), you get, b = 3 (you already know c, and d)
Now find a, which is 1.

Kiwi, chiplina chai bhayena ho. Chances are that you might get hurt too.
:)

There seems to be some problem with the notation in this site; instead of + and ? sign, I see question mark.
My first, second, and fourth equations (see the numbered ones) are all right. The third equation should be: 2 (a + b) = c ?(minus) 1.

Hawaguji,
>First Aa go to the other side of the river. Then, only A comes back leaving his gf alone on the other side. Then Bb take the boat and go to the other side. After Bb take off the boat, ...

Look here,
Mr. B will kiss Miss a in this stage.

Look the condition:
>Conditions: If a girl is in the different side that of her BF, she will be kissed by the other boys even if they are with their GFs.

Sorry, But hence your answer is not the right answer.

Nice to see so many math-aficionados here.

So here is one question I have always found intriguing. (Note that an answer to this question has been already given long b4 most of us here were born (u will be shocked to hear the answer) but it's such a delight to hear various (unique) opinions about this that i can't help posting this stale question once again.)

In our not-so-long life, we have solved or tried solving a lot of mathematical problems. For the sake of simplicity, a "mathematical problem" here will mean a problem in which we have to either prove that a certain "mathematically well-defined " statement is true or false (and nothing in between (no fuzzy here)) . And I assume there have been problems that after a lot of tinkering and contemplation still remain unsolved.

If the problem is one printed in a mathematics textbook , we suppose that it has a solution but we didn't have enough wit (one can blame his patience or lack of time too) to solve it. But what if a problem remains unsolved even when attempted by all great ppl from all around the world born at all times. (eg. GoldBach's conjecture - "Every even number greater than 3 can be expressed as a sum of 2 prime numbers": remained unsolved at least till one year ago). What can we infer about such problems : can we say that the problem is unsolvable ? or should we just settle it down by assuming that the problem is solvable but nobody has yet come up with a "brilliant" idea till now.

To generalize, here are two questions:
1) Can every statement be proved or disproved (disprove = to prove the converse ) mathematically ?
2) Is everything that can be proved mathematically true ?

Hoping for a nice discussion.

Guji,

Total length of the string = Circumference at P + Circumference at M
=2 *Pi *1+2*Pi*0.5=3Pi=3*3.1416 Cm^2

Sorry, Guji,
I forgot the linear distance.

Total length of the thread= 2*Average Circumferences of the 2 circles+ Linear Distance
= 2Pi(larger radius at P+smaller radius at M)+3
=2Pi(1+.5)+3
=3Pi+3
=3(Pi+1)
=3*3.1416
=9.4248 Square Cm

Sorry Guji,
I forgot the linear distance=3

The total length=3Pi+3=3(3.1416+1)=3*4.1416=12.4248 Square Cm

dammn this is fuN!! even though i havent got any answers right till now :P ..hmm like this thread..hey guji, thax for the answer on the other thread...

Sorry Bhinazu,
You are off by 56.54%. It does not look as simple as the way you do it. Have to think little more.
About the question you asked earlier, I thought that
[Conditions: If a girl is in the different side that of her BF, she will be kissed by the other boys even if they are with their GFs.]
meant that the guys cannot kiss the ladies if they are to go at the very same moment. Let me put you in more mathematical term: what I meant was at the instant when Bb get off the boat, a takes on the boat. So that the guy would not be able to kiss the lady. Think of it as limit..hehe

Anyway, if you really think that there is some other secured way, I can?t think on top of my head right now. Will post it later, if I come up with a better answer.

wrong. absolutely wrong. badmaash

Hehehe.. bhinazu's length of thread in sq cm...

Hey Hauguji ..good try...

I rather solve it in a easier way. Using tristate logic...
In binary there is 1 or 0. but in this trinary or tristate logic there are 3 stages ...-1( if the stone is in another pan) , 0 ( if the stone is not being used) and 1( if the stone is in the right pan while weighing)

since there are 4 stones and 3 logic levels.. there are 3 to the power 4 states(3^4 = 81)..( similar to that of binary where 2^4 states if 4 bits are used)
That gives us 81 states ... of which 40 are negative numbers(since there is -1 logic level), a zero and 40 positive numbers.
These positive 40 numbers are the 40 states or 40 combinations.

so now the weights of each stone ( similar to that of binary) wud be 3^0, 3^1, 3^2 and 3^3.
Thus the weights are 1, 3, 9 and 27.

Kiwi,
I do not have much knowledge about trinary, logics etc? although I can see what you are doing.

Question:
You need to find the value of x and y such that the equation below holds true.
x^2 - 11y! = 2003

(that?s y factorial)

Yea, it is
x = 45 and y=2
Thanks,

how I did is x^2 -11y! =2003 ..............eq (1)
or, x^2 =2003 +11y! and i took y =0 to go to nearest value so
x^2 =2003+11=2014 it gives x = 44.87
Again i took now X=45 and put the value in eq( 1) than I got y= 2

Have fun :)

is x an integer number? if not there are numerous solutions for this equation

it should be integer.

Next:
Question:
How do you write in number 11 hajaar 11 Saya 11. By using only 1s.

Oh yea.
Sorry for missing some info. x and y should be positive integer.

ricky, why thanks?? :)

Regarding the Boat question with Aa, Bb, Cc

Aa crosses the river on the boat, A comes back leaving a
Bb crosses the river, B comes back leaving b
Cc crosses the river, C comes back leaving c

Now,
A and B crosses the river, c comes back
and lastly
Cc goes across and lives happily ever after.

This one is easy one.

If there exists a positive real number x such that Cos[tan(inverse) (x)] = x, find the value of x^2 ( x square).