Answers to Base Conversion Exercises

 35
 738
 2393
 45871
 14040

 467_{8}; 137_{16}
 27365_{8}; 2EF5_{16}

 111100000011
 F03

 00111100100111100111 (the first two 0s aren't necessary!)
 744747

 1000010000_{2};
1020_{8}; 210_{16}; 1353_{7}
 1000010101101_{2};
10255_{8}; 10AD_{16}; 15306_{7}
 101010101_{2};
525_{8}; 155_{16}; 665_{7}

 2^{5}1 = 31
 2^{11}1 = 2047
 2^{31}1 = 2,147,483,647

 1111111100111010
 0000000000100101
 1111111111011011
 1110101110001101
 0010001110001110
 1101110001110010

 115
 119
 34
 7
 1

 The largest positive integer that fits in 12bit twos complement
is 2^{11}  1 = 2047 so 2548 is too big and overflow will occur.
In fact, the base 2 representation of 2548 is 100111110100 which is
12 bits but there needs to be at least one more bit for the sign. The
computer would take the 12 bits and interpret it as a 12bit twos
complement number. Hence it would report a value of 1548.
 The smallest integer that fits in 12bit twos complement is 2048
so 5377 is too small.