The Number of integral values of n (where $n ≥2$) such that the equation $2n\{x\} = 3x+2[x]$ has exactly 5 solutions (where [.] and [.] are respectively the greatest integer and fractional part functions), is ________.
Answer & explanation
We have,
$2n \{x\}=3x+2[x]$
$2n \{x\}=3([x]+\{x\})+2 [x]$
$⇒\{x\} =\frac{5[x]}{2n-3}$
$⇒0≤\frac{5[x]}{2n-3}<1$ $[∵0≤\{x\}<1]$
$⇒0≤[x]<\frac{2n-3}{5}$ $[∵n≥2]$
This will have five solutions [x] = 0, 1, 2, 3, 4, if $4<\frac{2n-3}{5}≤5⇒\frac{23}{2}<n≤14$ ⇒ n=12, 13, 14
Hence, n has 3 integral values.