A couple more questions on binary numbers?

Question:i'm really stuck.

what is the binary number used to represent the integer value of 57?

a. 0000 0057
b. 0011 1111
c. 0011 1001
d. 0110 0110

What is the binary number used to represent the integer value of 247?

a. 0111 1010
b. 1111 0111
c. 0000 0111

Answers:
Question #1: C
Question#2: B

I hope this helps and good luck with your homework!
Go to the website How Stuff Works. They explain the binary number system in detail. It will be under the computer section since they use it in programming.
thing you gotta remember is this;

Binary is a basic "yes or no" code.

Try doing a wikipieda search for binary...and do your own homework.

Good luck!
Me
I can't really explain this well but:
For the first question the answer is c.
That's because binary can be converted with the powers of two. From right to left, number powers of 2 from 2 to the power of 0 to two to the power of 7. Then, match the numbers that add up
128 64 32 16 8 4 2 1--> 0(because 128 is too big) 0 (again too big) 1 (32 could be added) 1 (32+16= 48) 1 (48+8=56) 0 (because 4 would make it too big) 0 (because 2 would make it too big) 1 (56+1=57).
1 represents that it should be added and 0 makes it null.

For 247, again follow the same process:
The answer is b.
128+64+32+16+4+2+1=247
Hi, I'll try to explain you the manual way to find out the answers.

For 57:
Divide 57 by 2 --> 57/2 = 28 with remaining 1 (28 x 2 = only 56, right...?)
Divide 28 by 2 --> 28/2 = 14 with remaining 0
Divide 14 by 2 --> 14/2 = 7 with remaining 0
Divide 7 by 2 --> 7/2 = 3 with remaining 1
Divide 3 by 2 --> 3/2 = 1 with remaining 1
Divide 1 by 2 --> 1/2 = 0 with remaining 1 (remember, you have to keep dividing until it is zero)

Now, list all the remaining we have:
1-0-0-1-1-1

Then, put it backward:
1-1-1-0-0-1

Notice that the multiple choices are all in 8 digits? So, just add up 2 zeros in front:
0011 1001

Bingo, you got your answer... =)

For 247, it's the same way, but since the number is bigger, it's definitely a bit longer...

Happy trying...!

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