Possible?

Relevant wiki: Adding Fractions

The sum of these unit fractions comes out to 86021/27720; multiply the equation by 27720, and the problem requires you to separate 12 integers into two groups each adding up to 86021/2=43010.5, not an integer, which is obviously impossible.

Intuitively, you can quickly figure out that if one side has 1/7, there's nothing on the other side which can "cancel it out", as 7 is coprime with every other dominator. Similarly with 1/11.

Relevant wiki: Adding Fractions

If this is possible, we can write $\frac 1{11} = \text{a linear, integer combination of the other fractions}.$ Now a linear combination (with integer coefficients) can only have denominators which divide the least common denominator of these other fractions; this least common denominator is $2^3\cdot 3^2\cdot 5\cdot 7$ .

However, since $11$ is not a divisor of this least common denominator, it cannot be expressed in this manner. Therefore the answer must be no.

Relevant wiki: Prime Numbers

Lets take the largest prime denominator $11$ .

Now let $L = \text{lcm}(1,2,3,4,5,6,7,8,9,10,12)$ note: $11$ is not present

If we multiply the equation by $L$ , we will have:

$(L \square \frac{L}{2} \square \frac{L}{3} \square ... \square \frac{L}{10} \square \frac{L}{12}) \square \frac{L}{11} = 0$

Because of how we created $L$ , all of the terms in $()$ will be integers, so any $\pm$ combination of them is also an integer:

$N \square \frac{L}{11} = 0$

But, since $11$ is a prime not included in $L$ , $11\nmid L$ , so $\frac{L}{11}$ is not an integer, and adding or subtracting it from any integer $N$ will not equal $0$ .

If i write each term under the same denominator (using the lowest common multiple for the denominator), each term is an even number except for 1/8 which will be the only even number. Hence no solution.

1/2 + 1/4 = 1/2, Similarly, 1/3+1/6= 1/5+1/10=1/2, 2 of these gets cancelled out. 1 - 1/2 (remaining) = 1/2. This 1/2 +1/12 = 7/12. 7/12 + 1/7+ 1/8+ 1/9 + 1/11 can in no way arranged to give a zero.

Amateur here: I found the LCD and wrote out all of the fractions like you will see: 13860/27720 🔲 9240/27720 🔲 through 2520/27720 🔲 2310/27720 = 0

At that point, as I was looking at the mathematical sentence, I realized that I couldn't possibly do anything to the final fraction, no matter what I did to all of the other ones, to get that last one to equal zero. (FYI: My initial impulse was to subtract the fractions, prior to realizing that I was never going to be able to zero out both sides.)

I know that my solution isn't theoretically based, extremely sophomoric, and it's totally ok if you laugh at me. However, I did get the correct answer and I wonder if there is room for thinking outside of the box.

A simple partity argument suffices, as well. Assume there is a solution. Multiply both sides of the equation by the LCM of the denominators, which is $2^3\cdot 3^2\cdot 5\cdot 7\cdot 11$ . Reducing mod 2 gives 1 on the left and 0 on the right, a contradiction.

By Riemann Series theorem, if an infinite series of real numbers is conditionally convergent, then its terms can be arranged in a permutation so that the new series converges to an arbitrary real number or diverges. More generally, p positive real numbers followed by q negatives gives the sum ln(p/q), which is not zero.

What you want to do is take one number and subtract several other terms to cancel it out, and repeat this until all terms are canceled out at 0. You can start with +1 - 1/2 - 1/3 - 1/6 = 0. However it becomes clear that you can't do anything to get rid of the 1/7 and 1/11 so the sum will always be a little off from 0

The easier way to the think about this problem is as two infinite series of progressive terms n/n, 1/(n+1), 1/(n+2). The highest value will be a sum all positive numbers which approaches n/n + n/n, or 2. The lowest value will be a sum of negative numbers approaching n/n - n/n or 0. All solutions must fall between these values, therefore no solution equals 0.