Shivam Pandya
1/11/18
QC2845
Homework_Assign#2
1. For decimal 35 to 8 bit will be 100011.
2. For decimal 32 to 8 bit 100000.
3. By using double add 00010101 to decimal is 21.
4. By using also double add 00011001 to decimal is 25.
5. We can identify by the remainder of the first division by two of the significant digit in binary number. If there is quotient and the remainder involves we can explain why LSD is odd or even.
6. Since we know in hexadecimal number you have to start counting 4 bit from right each, so for the last four bit 0101 binary I got 5 and four the first four 0001 I got 1, and if you combine all together you ended up with 15 bit.
7. Same with this question, lets start from last four bit 1001 binary I got is 9 and for the first four 0001 is 1, so you ended up with 19 total.
8. …show more content…
Lets find first hexadecimal for 0*15
0000001 1*16 = 16
0000101 5+16 = 21 +1
________
0000111
9. Lets first hexadecimal for 19
0000001 1*16 = 16
0001001 9 + 16 = 25
10. By looking the first question when we found 35 bit 100011, you just convert in 011101.
11. By looking the second question we found 32 bit, 100000, it will be remain same.
12. A) binary would be 01111111 which is 127, since it’s asking for two complement you just convert decimal which is -127.
B) Binary will be 00000001 which will be -1.
C) Binary should be 01010000 which the answer should be in two complement, -80 yet you don’t have to change anything since every thing is 0.
D) 00011111
E)
13. 128 +4 = 132 if you find the 2048
14. 2047 = 2 * 7^2 + 0 * 7^1 * 4 * 7^0 = 102
15. 2046 = same as top , 72 ( 36 *2 + 4 )
16. 2045 = ( 25 * 2 + 4 ) 54
17. 8110 =
22. 0 * 88 hexadecimal 64 *1 =