The example is 120 as well. And the subsets are:{120}-{60,30,15,10,5}-{40,24,20,12,8,6,4,3,2,1}. Now for the part b we're gonna use kinda induction. So suppose there exist a 3-stratum number x with sub sets A={a1,a2,...,ak},B and C. Now indeed there exist a prime number p that does not divide x , name (x*p) as y. We aver y is also a 3-stratum number since the subsets are:A2={p*a1,p*a2,....,p*ak,a1,a2,a3,....,ak}(B2 ans C2 are defined similarly!) So we are done! Many thanks to arashfree

Base case: $120$ is $3$-stratum with the sets $\{120\}, \{60, 40, 20\}, \{1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 24, 30\}.$ Inductive step: Let the original number be $x$. Multiply $x$ by $p$ such that $p$ is a prime and that $p\nmid x$. It is well known that the sum of the divisors of the number $n=p_1^{e_1}p_2^{e_2}\cdots$ is just \[ (1+p_1+p_1^2+\dots+p_1^{e_1})(1+p_2+p_2^2+\dots+p_2^{e_2})\dots \]So $\sigma(p\cdot x)$ is just $\sigma(x)\cdot (p+1)$. Now define the 3 sets of divisors of $x$ that have the same sum to be \[ A={a_1, a_2, \dots, a_{|A|}}, B={b_1, b_2, \dots, b_{|B|}}, C={c_1, c_2, \dots, c_{|C|}} \]Each of these sets has sum of $\frac{\sigma(x)}{3}$ and we want each of these sets to have a sum of $\frac{\sigma(x)\cdot (p+1)}{3}$. Now if we add the numbers $pa_1, pa_2, \dots, pa_{|A|}$ to set A and likewise for set B, then both of these sets would have a sum of $\frac{\sigma(x)\cdot (p+1)}{3}$. Set C could just be the leftover divisors of $px$ and it will have the same sum as the other two sets. For example if the starting number is $120$ (which we know is $3$-stratum) and multiply it by $7$ we get the number $840$. The sum of its divisors is $3 \cdot 960$ so we know that each set has to have a sum of $960$. Sets A, B are thus $\{840, 120\}, \{420, 280, 140, 60, 40, 20 \}$ and set C is just the leftover divisors of $840$.

Aran_mhdn wrote: The example is 120 as well. And the subsets are:{120}-{60,30,15,10,5}-{40,24,20,12,8,6,4,3,2,1}. Now for the part b we're gonna use kinda induction. So suppose there exist a 3-stratum number x with sub sets A={a1,a2,...,ak},B and C. Now indeed there exist a prime number p that does not divide x , name (x*p) as y. We aver y is also a 3-stratum number since the subsets are:A2={p*a1,p*a2,....,p*ak,a1,a2,a3,....,ak}(B2 ans C2 are defined similarly!) So we are done! Many thanks to arashfree How did you guess this number in part a? Thanks

congtuan wrote: How did you guess this number in part a? Thanks Hello, You can first try small numbers, but don't know if that gets anywhere . using prime factorization gives some ideas as well, it can be seen that if the number has 1 or 2 prime factors it can't be 3-stratum. After that if you try small numbers that have 3 prime factors, e.g. 2x3x5=30, 60, 90, 120, you'll find out that 120 is 3-stratum.