Suppose he initially had n posts. Assuming he had one post at each end, there area n-1 intervals between posts and the width of the property is 2(n-1). If he uses "3 posts less", then he uses n- 3 posts, n- 3- 1= n- 4 intervals between them and so the width of the property is 3(n- 4). Since both 2(n- 1) and 3(n- 4) give the width of the property, set them equal and solve that equation for n. Put that vaue of n back into either formula to get the width of the property.