How 127 is represented in binary should be detailed in the process

Divide by 2 to take the remainder, until the quotient is 0, read the remainder from the bottom up!For example, 127 2=63, take the remainder 1, then the last digit of the 127 binary is 1 63 2=31, take the remainder 1, then the penultimate digit of the 127 binary is 1 31 2=15, take the remainder 1, then the third penultimate digit of the 127 binary is 1 15 2=7, take the remainder 1, then the fourth penultimate digit of the 127 binary is 1 7 2=3, take the remainder 1, then the penultimate digit of the 127 binary is 1 3 2=1, take the remainder 1, then the penultimate digit of the 127 binary is 1 Finally, 1 1 = 1, then the highest digit of 127 is 1 Because it is an integer part is read from the bottom up, so the highest digit should be the last remainder 1If there is a decimal part, the decimal part is multiplied by 2 and rounded until the accuracy is integer, and the reading is from top to bottom.

The opposite of the integer part! You should be able to understand this way.

Conversion between decimal and binary numbers (1) Convert decimal integers to binary integers The way to convert a decimal integer to a binary integer is as follows: Divide the converted decimal integer by 2 repeatedly until the quotient is 0, and the resulting remainder (read from the last digit) is the binary representation of the number. To put it simply, yes"Divide by 2 remainder method".

For example, the method of converting a decimal integer ( 2 1 5 ) 1 0 to a binary integer is as follows: So, ( 2 1 5 ) 1 0 = ( 11 0 1 0 111 ) 2 is converted to a hexadecimal integer"Divide by 1 6 remainder".

127 to binary is 1111111 (binary).

Divide by 2 and take the remainder:

127 divided by 2, quotient 63, remainder.

Yes 1; 63 divided by 2, quotient 31, the remainder is 1;

31 divided by 2, quotient 15, the remainder is 1;

15 divided by 2, quotient 7, the remainder is 1;

7 divided by 2, the quotient is 3, and the remainder is 1;

3 divided by 2, quotient 1, the remainder is 1;

1 divided by 2, quotient 0, and the remainder is 1

Arrange the remainders from bottom to top, so convert 127 to binary to 1111111.

Answer: Decimal 221 converted to binary equals 11011101b.

The steps to solve the problem are as follows:

221÷2=110…The remainder 1,1 2 0=1,110 2=55....The remainder is 0,0 2 1=0,55 2=27....The remainder 1,1 2 2=4,27 2=13....The remainder 1,1 2 3=8,13 2=6....The remainder 1,1 2 4=16,6 2=3....The remainder 0,0 2 5=0,3 2=1....The remainder is 1, the year is 1 2 6 = 64, 1 2 = 0....The remainder 1,1 2 7=128, verification: 128 + 64 + 0 + 16 + 8 + 4 + 0 + 1 = 221, get: (221) hu car = (11011101).

Answer: The decimal grinding 126 is converted to binary equal to 0b1111110.

The steps to solve the problem are as follows:

126÷2=63…The remainder 0,0 2 0=0,63 2=31....The remainder 1,1 2 1=2,31 2=15....The remainder 1,1 2 2=4,15 2=7....The remainder 1,1 2 3=8,7 2=3....The remainder 1,1 2 4=16,3 2=1....The remainder 1,1 2 5=32,1 2=0....Remainder 1, 1 2 6 = 64, cave skin.

Validation: 64 + 32 + 16 + 8 + 4 + 2 + 0 = 126, gets: (126) Jane Trembling = (1111110).

Answer:Divide a decimal number by two, divide the obtained quotient by two, and so on until the quotient is equal to zero, and the remainder of the division is the converted binary result.

Answering process:127 2 = 63 remainder 1

63 2 = 31 remainder 1

31 2 = 15 surplus 1

15 2 = 7 surplus 1

7 2 = 3 remainder 1

3 2 = 1 and 1 remains

1 2 = 0 more than 1

Concatenating the remainder from bottom to top is the binary number that is sought: 127(10)=1111111(2).

Multiply by 2 and round up to 17 128*2=17 64....017/64*2=17/32...017/32*2=17/16(1/16...

1);1/16*2...The final answer is.

Binary. It is a numerical technique that is widely used in the operation of the spring species of one limb. Binary data is a wide base number represented by two numbers, 0 and 1.

17 can be paired with binary or expressed as () Zhijingwu.

Correct answer: D

127 is an odd number, and the last digit of binary is 1,127-1=126, and 126 is the exponential value to be broken down into 2: 2 6+2 5+2 4+2 3+2 +2

So it's 1111111

In the base binary, except for the last 1 represents 1, the 1 of each bit from the last digit to the left represents the progressive exponential value of 2, and 0 represents vacancy.

Divide by 2 consecutively, and then write the remainder continuously from bottom to top.