Consider P=(1+101001)10100,
Let x=10100
⇒P=(1+x1)x
⇒P=1+(x)(x1)+⌊2(x)(x−1)⋅x21+33(x)(x−1)(x−2)⋅x31+…
(upto 10100+1 terms)
⇒P=1+1+(⌊21−⌊2x1)+(⌊31−…)+… so on
⇒P=2+(Positive value less than ⌊21+⌊31+⌊41+… )
Also e=1+⌊11+⌊21+⌊31+⌊41+…
⇒⌊21+⌊31+⌊41+…=e−2
⇒P=2+ (Positive value less than e−2)
⇒P∈(2,3)
⇒ Least integer value of P is 3.