Han found a way to compute complicated expressions more easily. Since 2 β
5 = 10 2β
5=10β, he looks for pairings of 2 2βs and 5 5βs that he knows equal 10 10β. For example, 3 β
2 4 β
5 5 = 3 β
2 4 β
5 4 β
5 = ( 3 β
5 ) β
( 2 β
5 ) 4 = 15 β
1 0 4 = 150 , 000 3β
2 4 β
5 5 =3β
2 4 β
5 4 β
5=(3β
5)β
(2β
5) 4 =15β
10 4 =150,000 β Use Han's technique to compute the following: a) 2 4 β
5 β
( 3 β
5 ) 3 2 4 β
5β
(3β
5) 3
