CAT
Exam
If n is an odd integer then the last two digits in base ten of 2^2n * (2^(2n+1) -1) are
I. 18 II. 28 c. 38
a. Only I
b. Only I & II
c. Only II
d. Only I, II & III
e. None of the above
Read Solution (Total 1)
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- c. Only II
when n=1
2^2n * (2^(2n+1) -1)= 4*(8-1)=4*7=28
when n=3
2^2n * (2^(2n+1) -1)= 64*127=8128 - 12 years agoHelpfull: Yes(1) No(1)
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