One can first try to compose the sum of 1 with smaller number of reciprocals. Let's start with smallest positive integers. 1/2 + 1/3 < 1, but 1/2 + 1/3 + 1/4 > 1, and the part 1 - (1/2 + 1/3) is actually 1/6. So, with 3 positive integers we can have 1/2 + 1/3 + 1/6 = 1.
Now we can try to replace 1/6 with smaller reciprocals, and we need exactly 3 of them, so the problem sort of repeats itself - now we need to find 3 such integers, that their reciprocals sum to 1/6. This is the same problem as we just solved, only scaled 1/6 times, so instead of 1/2, 1/3 and 1/6 we can have them divided by 6 - so, 1/12, 1/18 and 1/36.
So now we have five numbers - 1/2, 1/3, 1/12, 1/18 and 1/36 - summing to 1 and their reciprocals are integers.
You suspect that's not the only solution - after all, original 1/2 + 1/3... was selected rather arbitrarily, as the largest numbers with reciprocals still less than 1.
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[ 2.9 ms ] story [ 22.3 ms ] threadNow we can try to replace 1/6 with smaller reciprocals, and we need exactly 3 of them, so the problem sort of repeats itself - now we need to find 3 such integers, that their reciprocals sum to 1/6. This is the same problem as we just solved, only scaled 1/6 times, so instead of 1/2, 1/3 and 1/6 we can have them divided by 6 - so, 1/12, 1/18 and 1/36.
So now we have five numbers - 1/2, 1/3, 1/12, 1/18 and 1/36 - summing to 1 and their reciprocals are integers.
You suspect that's not the only solution - after all, original 1/2 + 1/3... was selected rather arbitrarily, as the largest numbers with reciprocals still less than 1.
Well, I know this:
We're trying to get to 5 terms. So let's split 1/2 into two terms. Hmm, let's do that again - this time splitting 1/4. Ok, I'm pretty close. Can I split 1/8 into 1/9 and some mystery term? Well, that's not hard to solve. N = 72.1 = 1/2 + 1/4 + 1/8 + 1/9 + 1/72