會考數學一問,急.......會考數學一問,急.......

110 110 111 110 001 (2)
= 2^14 + 2^13 + 2^11 + 2^10 + 2^8 + 2^7 + 2^6 + 2^5 + 2^4 + 1
= 4*16^3 + 2*16^3 + 8*16^2 + 4*16^2 + 16^2 + 8*16 + 4*16 + 2*16 + 16 + 1
= 6*16^3 + 13*16^2 + 15*16 + 1
= 6DF1(16)
(D)

2010-04-25 02:30:59 補充：
方法2 :

110 110 111 110 001 (2)

= 2^14 + 2^13 + (2^11 + 2^10 + 2^8) + ( 2^7 + 2^6 + 2^5 + 2^4) + 1

= (2^2 + 2^1)(2^12) + (2^3 + 2^2 + 2^0)(2^8) + (2^3 + 2^2 + 2^1 + 2^0)(2^4) + 1

= (6)(16^3) + (13)(16^2) + (15)(16^1) + 1

= 6DF1(16)

(D)

Actually, you can look at this in this way . Since 110110111110001(2) is in binary, right? If you want to change it to 十六進制, what you have to do is:

1.) separate it into 4 numbers a group

110 1101 1111 0001

(1) (2) (3) (4)


Because the first group (1) only got 3 numbers, we will give it one more zero in the beginning, so it will become

0110 1101 1111 0001


You should have learned the conversion:

Binary(2 進制) Hex (十六進制)
0001 1
0010 2
0011 3
0100 4
0101 5
0110 6
0111 7
1000 8
1001 9
1010 A
1011 B
1100 C
1101 D
1110 E
1111 F


So right now, we will convert the 4 groups of numbers


0110 1101 1111 0001

0110 -> 6
1101 -> D
1111 -> F
0001 -> 1

So the answer is D. If you got anything you don't understand, tell me. Wish 會考 successful ! ^^

2010-04-25 03:37:54 補充：
I hope The conversion table is not too messy for you to understand:
Binary(2 進制) Hex (十六進制)
0001 1
0010 2
...... etc
1010 A
1011 B
........ etc

It means 0001(2) is 1 in 十六進制
1010(2) is A in 十六進制