Question 363785
If he leaves home with x cakes, then by the times he leaves the 1st bridge, he has {{{x/2 + 1}}} cakes.  When he leaves the 2nd bridge, he has {{{(1/2)(x/2+1)+1 = x/2^2 + 1/2 + 1}}} cakes.  When he leaves the 3nd bridge, he has {{{(1/2)((1/2)(x/2+1)+1)+1 = (1/2)(x/2^2 + 1/2 + 1)+1 = x/2^3 + 1/2^2+1/2 +1}}} cakes.  If there are n bridges, then inductively when he leaves the nth bridge, he must have {{{ x/2^n + 1/2^(n-1)+1/2^(n-2)}}}+...+{{{1/2^2+1/2 + 1}}} cakes.  this must be equal to 2.  Hence
{{{ x/2^n + 1/2^(n-1)+1/2^(n-2)}}}+...+{{{1/2^2+1/2 + 1 = 2}}}.
{{{x/2^n + (1-1/2^n)/(1/2) = 2}}},
{{{x/2^n + 2 - 1/2^(n-1) = 2}}},
{{{x/2^n - 1/2^(n-1) = 0}}},
{{{x/2^n = 1/2^(n-1)}}}, 
{{{x = 2^n/2^(n-1)}}}, or x=2.