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 ________.

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.