casebash said:
For qu 30, one vital fact is if two does not divide your number your number is the sum of 4 odds which is an even and contradiction. I got that off Graham White who solved all the questions.
And the rest of the 'proof' is thus:
Let the number be n
We know that two of the 4 smallest divisors will be 1 and 2.
If another is 3, then we have n = 1 + 4 + 9 + a^2 = 14 + a^2. a/n, so a/14 (as a/a^2). We check 7 and 14, and they don't work.
Therefore 3 does not divide n
We check 4 in much the same way. (n = 21 + a^2, a/21, a = 1,3,7 or 21, but none work)
Therefore 4 does not divide n
We check 5, and get n = 30 + a^2. Checking the divisors of 30, we note that a = 10 works. This gives 130, which has a largest prime divisor of 13.
We have not proven that this (130) is the only number to satisfy these conditions (this would be an interesting question) but the question implies that there is only one (or at least that all of them have the same largest prime divisor). Therefore the answer is 13.
