Why is the sum of positive integers equal to -1/12?
99187
This was a form of math that gained a small following several years ago with the Hmong of Thiland. It's not used very much today due to its limited application.   Edited comment:  Pastor Dave is correct, it's still used. I think it was also used for a few months at Enron, Corp.
734075
6(1/2)=3\n-2(1/2)=-1\n12(1/2)=6
948937
(-3,-3);(-2,0);(-1,1);\n(0,0);(1,-3)
944997
you divide or multiply by a negative numberand the sign of inequality flips
395828
haha, but this does not work. you cannot mathematically divide by (a-b) because this quantity is equal to 0. (1-1)=0. Dividing by zero yields an indeterminate answer.
1052440
The trivial zeros are at (-2K+0i), where K is a positive integer. The nontrivial zeros are all the others. It has been shown that the nontrivial zeroes must have a real part between 0 and 1.
1043253
A negative number may be an integer, but not a whole number.\n\nWhole numbers are "counting numbers" starting with zero: 0, 1, 2, 3, 4, . . .\nIntegers are everything without a decimal point: . . . -3, -2, -1, 0, 1, 2, 3, . . .\n1934234 is a whole number and an integer\n-36 is an integer, but not a whole number\n27.5 and -3.12 are neither whole numbers nor integers
501645
I too failed math but to me this is a logic question:\n0+0=0\nWhat else cloud it be?
457195
the simple program given above if fine.. in terms of the conncept.. but there's a bug... \nthe problem is with looping.. coz when u enter a number such as 80301  or 55302 ( a no. having 0 in between), the loop breaks at the point when zero is encountered.\n\n so the output in the first eg. is  8 0 ,sum= 8 (instead of correct 8 0 3 0 1 , sum=12\n\n\nso the logic of handling the number is absolutely correct but the looping is a comparatively difficult job in this case!
8564
It looks to me that the integer part means that you take the integer that comes before the given number if the given number is not an integer thus 0 would be the correct answer.
144204
-10\n\nand to prove it -10 x -10 x -10 = -1000
607947
i^-1= 1/i^1  you can get rid of the negative exponent by changing it to the reciprocal.
667025
You should verify this.\n_0_2_4_6_8_10_12\n0|0 0 0 0 0  0  0\n2|0 4 8 12 2 6 10\n4|0 8 2 10 4 12 6\n6|0 12 10 8 6 4 2\n8|0  2  4 6 8 10 12\n10|0 6 12 4 10 2 8\n12|0 10 6 2 12 8 4
89444
The answer is "i" because there is no number, that when mutliplied by itself, will give you -1.  Think about it: the only possible combination that would make sense is 1 X 1, which equals 1...not -1.
680703
Integers are all of the natural numbers.Are you asking about negative and positive integers and why are the signs changed at times?\nIf that is the case,you cannot subtract a negative from a negative because you get a positve.It is counter-intuitve but if you think about it,it makes sense.\nFor instance a -6--6=what?It all depends upon where the signs are placed and if it a parenthetical(if it has parentheses) expression.\nDo you see how that problem can have a number of solutions,depending upon how it is expressed(set up)?\nThe answer is concievably a -12 or it can be zero.\nYou must first understand that the answer may not be evident.\nThere is no easier way to deal with it.You must understand it in your mind:it is strictly an intellect thing.\nIt really is a mind game in ways,but it is a good one.You can do it,it's fun in ways.It teaches you to think and to widen the mind.\nThere is a method to the madness however and once you get that straightend out in your mind,you will feel like a genius.\nIt always works the same and it never works differently:all you have to do is look for the sense of it and ask questions.\nIn the example that I gave,a negative six subtracted from a negative six is a negtive 12.However a double like sign with a negative number cancels itself out and becomes positive.The answer is zero,because there are no parentheses.
613146
Nice problem. Here's the solution: 1000! is a multiple of 9 which is the same thing with saying that: S(1) = the sum of its digits is a multiple of 9 (Using the following known fact: a number is divisible by 9 if and only if the sum of its digits is divisible by 9). \n   Clearly S(1) < 1000! and 9 | S(1). Summing the digits of S(1) one obtains a number S(2) which is strictly smaller than S(1) and is a multiple of 9. And so on, until finally one arrives at a one digit number, say S(r) for some r  which is divisible by 9. But this can only be 0 or 9. Since S(r) is the sum of the digits of the previous number obtained by this algorithm, namely S(r-1), it can not be 0. So it must be 9! \n   Note that this type of proof works for any number divisible by 9, not only factorials of numbers, and in fact, with exactly the arguments displayed above it can be prooved that the single digit after adding the digits of an arbitrarly chosen number N is the digit d > 0 such that N-d is a number divisible by 9. (If you don't belive it, simply work out a computer programme that verifies this for all the numbers smaller than 1 000 000 for example).
420852
Because (no+1) is like adding +1 and -1, in other words the sum is zero.  When you cancel out the (no+1) from both sides in line 4, you are dividing by zero, which gives rise to all kinds of contradictions and paradoxes in mathematical proofs, even proofs like this one.