lf n > 0 and exactly 15 integers satisfy (x+6)(x−4)(x−5)(2x−n)≤0 then sum of digits of the least possible value of n is
10
12
14
16
(x+6)(x−4)(x−5)(2x−n)≤0
Hence, x∈[−6,4]∪5,n2
Now, there are 11 integers in [-6, 4].
So, there must be 4integers in 5,n2
So, n2∈[8,9)
⇒ n∈[16,18)