Question 136627
He starts with x dollars, so in the first store he spends {{{x/2+1}}}.  So when he leaves the first store he has {{{x-(x/2+1)=x/2-1=x/2-2/2=(x-2)/2}}}.


So in the second store he spends {{{((x-2)/(2*2))+1}}} so he has {{{((x-2)/(2*2))-1=(x-2-4)/4=(x-6)/4}}} left.


And so on.


I'll leave it to you to verify that when he leaves the 5th store he has {{{(x-62)/64}}} dollars remaining, and by the conditions of the problem this expression must be equal to zero (because he spent all of his money).  If  {{{(x-62)/64=0}}}, then {{{x-62=0}}} => {{{x=62}}}


Check:
Start: 62
Half of 62 is 31 plus 1 is 32, leaving 30
Half of 30 is 15 plus 1 is 16, leaving 14
Half of 14 is 7 plus 1 is 8, leaving 6
Half of 6 is 3 plus 1 is 4, leaving 2
Half of 2 is 1 plus 1 is 2, leaving 0