for an integer $\mathscr k$ ($2\leq k\lt500$) four positive integer a,b,c,d satisfies following conditions
- $2\leq a,b,c,d\leq k$
- $a^{\frac{1}{b}}\times c^{\frac{1}{d}}=24^{\frac{1}{5}}$
for k which makes $\vert{\{(a,b,c,d)\}}\vert=59$, find maximum and minimum value of k.
($\vert{\{(a,b,c,d)\}}\vert$ denotes the number of (a,b,c,d)
I tried brute force approach and obtained
k= 6: (4 5 6 5) (6 5 4 5) 2pairs
k= 8: (3 5 8 5) (8 5 3 5) 2pairs
k= 10: (9 10 8 5) (8 5 9 10) 2pairs
k= 12: (2 5 12 5) (12 5 2 5) (4 10 12 5) (12 5 4 10) 4pairs
k= 15: (8 15 12 5) (12 5 8 15) 2pairs
k= 16: (16 10 6 5) (6 5 16 10) 2pairs
k= 18: (18 10 2 2) (2 2 18 10) (18 10 4 4) (4 4 18 10) (18 10 8 6) (8 6 18 10) (18 10 16 8) (16 8 18 10) 8pairs
k= 20: (16 20 12 5) (12 5 16 20) 2pairs
k= 24: (24 10 24 10) 1pair
k= 27: (27 15 8 5) (8 5 27 15) 2pairs
k= 30: (24 6 24 30) (24 30 24 6) 2pairs
k= 32: (32 25 12 5) (12 5 32 25) (18 10 32 10) (32 10 18 10) 4pairs
k= 36: (4 5 36 10) (36 10 4 5) (16 10 36 10) (36 10 16 10) 4pairs
k= 48: (18 35 48 7) (48 7 18 35) (12 10 48 10) (48 10 12 10) 4pairs
k= 64: (27 15 64 10) (64 10 27 15) (9 10 64 10) (64 10 9 10) (18 10 64 15) (64 15 18 10) (3 5 64 10) (64 10 3 5) (6 5 64 15) (64 15 6 5) (64 15 36 10) (36 10 64 15) (64 30 12 5) (12 5 64 30) 14pairs
k= 72: (8 10 72 10) (72 10 8 10) (64 20 72 10) (72 10 64 20) 4pairs
summing all gives 59 pairs. and if k=81, there exists (81 20 8 5) (8 5 81 20) (81 20 64 10) (64 10 81 20) so $72\leq k\leq80$
But the answer was 359 (maxima+minima) and I don't understand why. This is a question in an important exam in my country. Please help me.