23. Convert the following numbers from the base shown to base 10. a. 111 (base 2

Question

23. Convert the following numbers from the base shown to base 10.
a. 111 (base 2)
b. 777 (base 8)
c. FEC (base 16)
d. 777 (base 16)
e. 111 (base 8)
28 (d) Convert the following binary numbers to octal:
1100010
29 (a) Convert the following binary numbers to hexadecimal.
10101001
(c) Convert the following binary numbers to hexadecimal.
01101110
32 (c ) Convert the following decimal numbers to octal:
1492
(e) Convert the following decimal numbers to octal:
2001
33 (d) Convert the following decimal numbers to binary:
99
34 (e) Convert the following decimal numbers to hexadecimal:
43

Explanation / Answer

23 a. 7 b. 7*64+7*8+7*1 = 511 c. 15*16^2+14*16+12*1 = 4076 d. 1*64+1*8+1*1 = 73 28 d. 0b1100010 = 0b001100010 Break it into groups of 3 001 100 010 Convert each one into decimal 1 4 2 Answer is 142 in octal 29 a. Break it into groups of 4 bits 1010 1001 Convert each group to decimal 10 9 Convert each group to hexadecimal A9 c. 0110 1110 6 14 6E 32 c. Octal has a 1's place, 8's place, 64's place, 512's place, 4096's place, and so on. We know that 1492 is less than 4096 but greater than 512. So, let's divide by 512 and find the remainder 1492 / 512 = 2 r 468 Now divide the remainder by 64 and repeat 468 / 64 = 7 r 20 20 / 8 = 2 r 4 So, our answer is 2724 e. 2001 / 512 = 3 r 465 465 / 64 = 7 r 17 17 / 8 = 2 r 1 3721 is the answer 33 d. Recall that bits in binary go 1, 2, 4, 8, 16, 32, 64, 128, .... We use a similar approach to above. 99 is greater than 64 but less than 128 99 / 64 = 1 r 35 35 / 32 = 1 r 3 skip 16 skip 8 skip 4 3 / 2 = 1 r 1 1100011 34 e. 43 / 16 = 2 r 11 11 is B in hexadecimal So the answer is 2B

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