I quickly came up with three solutions:
a) k_1= 1, n=1
b) k_1 = .. = k_4 = 1, n=4
c) k_1 = 2, k_2 = 3, k_3 = 6, n=3
Let's try to constrain the k. Suppose k_1 >= k_2 >= .. >= k_n then
k_1 <= (5n-4) -(n-1)*1 = 4n-3. Also
n/k_n >= 1/k_1 + .. + 1/k_n = 1 >= n/k_1, so
k_1 >= n >= k_n
Edited on May 1, 2020, 7:36 pm
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Posted by FrankM
on 2020-04-26 23:15:06 |