The person has no toffees remaining after distributing to the fifth student.
Each student receives one more than half the number of toffees available at that point.
For this problem, working backward is the most efficient approach. If the person had not given an additional toffee, they would have retained that single toffee.
This implies the person had \(2\) toffees at that stage. In the preceding stage (the 4th stage), they must have had \((2+1)×2\), which equals \(6\) toffees. In the 3rd stage, they should have had \((6+1)×2\), totaling \(14\) toffees.
In the 2nd stage, they should have had \((14+1)×2\), resulting in \(30\) toffees.
In the 1st stage, they should have had \((30+1)×2\), which amounts to \(62\) toffees.
Therefore, the initial number of toffees was \(62\).