You could create a molar solution rather than a dilution to prepare the 0.1M solution of HCl. By definition, molarity is moles of solute per liter of solvent. Therefore, to create a .1 M solution of hydrochloric acid, you would essentially have .1 moles of HCl per one mole solvent (which is usually water).
To make a liter of simple molar solution from a dry reagent, you would multiply the desired molarity by the formula weight to determine how many grams to use. So, for HCL in this question, this would be:
36.51 g/mole x .1 moles /L = 3.65 g / L
In other words, you would add 1 liter of water to 3.65 grams of HCl.
I hope this helps! And, as always, practice safety in the lab!