Let after k−1 guests have had a slice there is Xk-1cake left.
The kth guest gets Xk-1*k/100 cake.
The (k+1)th guest gets Xk-1(1−k/100)*k+1/100 cake.
Let us assume that the (k+1)th guest gets more cake than the kth guest, i.e.:
(1−k/100)k+1/100>k/100
(100−k)(k+1)>100k
−k2−k+100>0
The above equation is true for k=1…9, and it fails for for k≥10, so the 11th guest gets less than the 110th guest, who gets more than the 9th which implies that 10th guest gets the largest share of cake.
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