Is it possible to partition all positive divisors of 100! (including 1 and 100!) into 2 subsets so that each subset has the same number of integers and the product of all the divisors making up the first subset is equal to all divisors making up the second subset?
I have been trying to answer this problem using trial and error and have still not gotten to an answer. I was wondering whether there is an alternative method to simply trial and error. Additionally, I am also still finding it a bit hard to comprehend what the question is asking. I would really appreciate it if you explained to me how to do this question in an easy-to-understand manner. Thank you.
$100!=$
$2^{50+25+12+6+3+1}\cdot3^{33+16+8+4+2+1}\cdot5^{20+10+5+2+1}\cdot7^{14+7+3+1}\cdot$
$11^{9+4+2+1}\cdot13^{7+3+1}\cdot17^{5+2+1}\cdot19^{5+2+1}\cdot23^{4+2+1}\cdot$
$29^{3+1}\cdot31^{3+1}\cdot37^{2+1}\cdot41^{2+1}\cdot43^{2+1}\cdot47^{2+1}\cdot$
$53^1\cdot59^1\cdot61^1\cdot67^1\cdot71^1\cdot73^1\cdot79^1\cdot83^1\cdot89^1\cdot97^1$
Since $100!$ has a prime factor that appears an odd number of times, it is not a perfect square.
Hence the number of divisors of $100!$ is even, and every divisor $d$ can be paired with divisor $\frac{100!}{d}$.
Moreover, the number of divisors of $100!$ can be easily calculated as:
$98\cdot65\cdot39\cdot26\cdot$
$17\cdot12\cdot9\cdot9\cdot8\cdot$
$5\cdot5\cdot4\cdot4\cdot4\cdot4\cdot$
$2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2=k$
So define subset #$1$ as $\{d_i|i\in[1,\frac{k}{4}]\cup[k-\frac{k}{4}+1,k]\}$ and subset #$2$ as $\{d_i|i\in[\frac{k}{4}+1,k-\frac{k}{4}]\}$.