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$(110101101)_2=  (0001 \ 1010 \ 1101)_H$ (making group of 4 bits as in hexadecimal representation $\log_2 {16}=4$ bits.

$(0001)_2 = (1)_{10}=(1)_H$

$(1010)_2 = (10)_{10}= (A)_H$

$(1101)_2 = (13)_{10}= (D)_H $

Answer is: $(1AD)_H$
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