How to open Outlook 2016 with vs2017 - vb.net

Question

I've on my computer Office 2013 32bits. I develop an Intranet but on server and windows session of my customer : it's an Office 2016 32bits...
On my vs2017 I install with nugget package microsoft.office.interop 15.0.4797.1003 But it's for Office 2013.
I create mail with
dim outl as new outlook.application
dim Mai las outl.mailitem=outl.createItem(Micrososft.office.Interop.outlook.olitemtype.olmailItem)
Mail.To=...
...
Mail.attachment.add(MyFile)
Mail.display()
I try to put my code on the customer server, and like I thougth ... 80070005 Access denied...
I don't find how to open an Outlook 2016
Thanks for your help
UPDATE
On Property of my project, I've "Any CPU" option activate

Answer 1

Finally...
A friend says me, my method can be ok only If I open session on server. outlook.display run on server not on clientside on other computer.
Then I change my mind and build a emml file, store and push with ashx file
For Each _file As String In IO.Directory.GetFiles(IO.Path.GetDirectoryName(Path), "*.eml")
IO.File.Delete(_file)
Next
Dim MailMdp As New Net.Mail.MailMessage
MailMdp.Subject = Vers.Devi.Libelle
MailMdp.From = New Net.Mail.MailAddress(ConnectedUser.Mail)
MailMdp.To.Add(AnAdress)
MailMdp.Body = "Bonjour,"
Dim att As New Net.Mail.Attachment(Path)
MailMdp.Attachments.Add(att)
MailMdp.IsBodyHtml = True
Dim Client As Net.Mail.SmtpClient = New Net.Mail.SmtpClient()
Client.DeliveryMethod = Net.Mail.SmtpDeliveryMethod.SpecifiedPickupDirectory
Client.PickupDirectoryLocation = IO.Path.GetDirectoryName(Path)
Client.Send(MailMdp)
Dim MonEml = IO.Directory.GetFiles(IO.Path.GetDirectoryName(Path), "*.eml", IO.SearchOption.TopDirectoryOnly).First
Dim fso As New System.IO.FileInfo(MonEml)
Dim NomFichier As String = fso.Name
context.Response.Clear()
context.Response.ClearContent()
context.Response.ClearHeaders()
context.Response.ContentType = "message/rfc822"
context.Response.AppendHeader("Content-Disposition", "attachment; filename=" & NomFichier)
context.Response.TransmitFile(MonEml)
context.Response.End()
I hope that can be helpfull