Convert 34947ABCD.EF123993 to a decimal, octal and binary number. Also, check if this is a binary number?

Question

is it possible?

Answer 1

34947ABCD.EF123993 hex (base 16)



D*16^0 = 13*1 = 13 decimal

C*16^1 = 12*16 = 192

B*16^2 = 11*256 = 2,816

A*16^3 = 10*4096 = 40,960

7*16^4 = 7*65536 = 458,752

4*16^5 = 4*1048576 = 4,194,304

9*16^6 = 9*16777216 = 150,994,944

4*16^7 = 4*268435456 = 1,073,741,824

3*16^8 = 3*4294967296 = 12,884,901,888



E/16^1 = 14/16 = 0.875 decimal

F/16^2 = 15/256 = 0.05859375

1/16^3 = 1/4096 = 0.000244140625

2/16^4 = 2/65536 = 0.000030517578125

3/16^5 = 3/1048576 = 0.00000286102294921875

9/16^6 = 9/16777216 = 0.000000536441802978515625

9/16^7 = 9/268435456 = 0.0000000335276126861572265625

3/16^8 = 3/4294967296 = 0.00000000069849193096160888671875



Add them all up to get 14,114,335,693.93387183989398181438446044921875 decimal.



To convert to octal, start by finding the largest "factor of 8" that will be less than 14,114,335,693.



8^11 = 8,589,934,592 (which is lower) and 8^12 = 68,719,476,736 (which is too high) so we start with 8^11 (12 octal digits).



2 * 8^11 = 17,179,869,184 > 14,114,335,693 so the first digital must be 1. 14,114,335,693 - 1 * 8^11 = 5,524,401,101 remaining.



8^10 = 1,073,741,824 and 5*8^10 = 5,368,709,120 so the second digit is 5. 5,524,401,101 - 5*8^10 = 155,691,981 remaining.



8^9 = 134,217,728 so the third digit is 1. 155,691,981 - 1*8^9 = 21,474,253 remaining.



8^8 = 16,777,216 so the forth digit is 1. 21,474,253 - 1*8^8 = 4,697,037 remaining.



8^7 = 2,097,152 so the fifth digit is 2. 4,697,037 - 2*8^7 = 502,733 remaining.



8^6 = 262,144 so the sixth digit is 1. 502,733 - 1*8^6 = 240,589 remaining.



8^5 = 32,768 and 7*8^5 = 229376, so the seventh digit is 7. 240,589 - 7*8^5 = 11,213 remaining.



8^4 = 4096 and 2*8^4 = 8,192 so the eighth digit is 2. 11,213 - 2*8^4 = 3,021 remaining.



8^3 = 512 and 5*8^3 = 2560 so the ninth digit is 5. 3,021 - 5*8^3 = 461 remaining.



8^2 = 64 and 7*8^2 = 448 so the tenth digit is 7. 461 - 7*8^2 = 13 remaining.



8^1 = 8 so the eleventh digit is 8. 13 - 8 = 5 remaining.



8^0 = 1 so the last digit is the number 5.



So far we have 14,114,335,693 decimal = 151121725715 octal. The same must be done with the fractional part of the decimal number (.93387183989398181438446044921875).



The binary (base 2) conversion can be done the same way.



BTW - The scientific calculator that is built into Windows can do the integer number conversions to/from hex/octal/decimal/binary in the blink of an eye.



This should get you started. Enjoy.

Answer 2

===First, this is a HEX number based on the digits used.



===Hex to Bin conversion



Each individual Hex digit will become a four digit Bin number (or 'nibble'). The placeholders in BIN are based on exponents of 2, since we're working with Base2. So, from the right, the placeholders are 2^0 (2 raised to 0), 2^1, 2^2, and 2^3. This means, from the right, that the placeholders are 1,2,4, and 8.



You can reverse that to read a bin value from left to right

8,4,2,1



So a BIN of 0101 equals 4+1 or 5.

0011 = 2+1 or 3

and so on.



Finally 0000=0 and 1111=15, so that gives us the ability to represent 16 digits (0-15) and easily convert Hex to Bin.



= Your HEX number, 34947ABCD.EF123993

Starting from the left,

3 = 0011

4 = 0101

9 = 1001

4 = 0100

7 = 0111

A (10) =1010

B (11) = 1011

C (12) = 1100

D (13) = 1101

E (14) = 1110

F (15) = 1111

1 = 0001

2 = 0010

3 = 0011

9 = 1001

9 = 1001

3 = 0011



You can string those from right to left for your binary value. Yahoo would hide some of the string if I tried to put it all together. I've added spaces so you can see it all.



0011 0100 1001 0100 0111 1010 1011 1100 1101 . 1110 1111 0001 0010 0011 1001 1001 0011



I've left the decimal point intact. I'm not sure what you're actually doing with that since fractional or decimal notation isn't normally used in bin, oct, or hex. It could be used as a separation of values, though.



===Octal

I find it's best to use the Binary number you just created to get your Octal value. First, we need to get it into a format for Base8 conversion. That's simply done by taking your binary string and making triplets instead of quartets of numbers.



For instance,

1011 becomes 1 011

pad with leading zeros,



The placeholders are the same 2^0, 2^1, and 2^2. We don't have a 2^3 for Base8.



That becomes 1,2,4



So

001 = 1

010 = 2

111 = 7



Giving us 0-7, or eight numbers.



Take the long binary string, and group the digits into groups of three starting from the right.



Your Bin number is (with spaces so you can see it),

0011 0100 1001 0100 0111 1010 1011 1100 1101 . 1110 1111 0001 0010 0011 1001 1001 0011



and becomes (with spaces, so you can see it),

001 101 001 001 010 001 111 010 101 111 001 101 . 11 101 111 000 100 100 011 100 110 010 011



The first value to the right of the decimal point only has two digits, so that can be padded with a leading zero (11 becomes 011) to fill out to three digits.



Following the same method of conversion,

001 = 1

101 = 5

001 = 1

001 = 1

010 = 2

001 = 1

111 = 7

010 = 2

101 = 5

111 = 7

001 = 1

101 = 5

011 = 3

101 = 5

111 = 7

000 = 0

100 = 4

100 = 4

011 = 3

100 = 4

110 = 6

010 = 2

011 = 3



Which looks like,



151121725715.35704434623





===Check

Using the left and right parts of your number



34947ABCD (hex) =

0011 0100 1001 0100 0111 1010 1011 1100 1101 (bin) =

151121725715 (oct)



EF123993 (hex) =

1110 1111 0001 0010 0011 1001 1001 0011 (bin) =

35704434323 (oct)



=== Decimal

The easiest, of course, is to use Calc. Set it to "View - Scientific" and convert any of the above Hex, Bin, or Oct values to Decimal.



To convert to Decimal manually, it's usually easiest to convert a BIN to HEX, then use Base16 placeholders, like this,



Using your HEX number, 34947ABCD.EF123993



3 * 16^8 = 12884901888

4 * 16^7 = 1073741824

9 * 16^6 = 150994944

4 * 16^5 = 4194304

7 * 16^4 = 458752

A (10) * 16^3 = 40960

B (11) * 16^2 = 2816

C (12) * 16^1 = 192

D (13) * 16^0 = 13



= 14114335693



.



E (14) * 16^7 = 3758096384

F (15) * 16^6 = 251658240

1 * 16^5 = 1048576

2 * 16^4 = 131072

3 * 16^3 = 12288

9 * 16^2 = 2304

9 * 16^1 = 144

3 * 16^0 = 3



= 4010949011



14114335693.4010949011



==========

Finally,



34947ABCD . EF123993 (hex) =



0011 0100 1001 0100 0111 1010 1011 1100 1101 . 1110 1111 0001 0010 0011 1001 1001 0011 (bin) =



151121725715 . 35704434323 (oct)



14114335693 . 4010949011 (dec)

Answer 3

This should be a hexadecimal number since it contains number greater than 2(impossible to be a binary number) and most F(equivalent to 15 in hexadecimal).

34947ABCD.EF123993 = 1.41143*10^10 = 1.1010010010100011101*2^33 = 1.511216 * 8^11

Answer 4

110100100101000111010101111001101,

11101111000100100011100110010011



Binary number is stated above relting hexadecimal number stated by you. Use a 32-digits calculator, which is usually available on a PC and separately enter 'whole number part' and 'decimal part' in "hexadecimal mode". Click on binary and "then merge both into a-value" you will get dead accurate answer !



Similarly Octal is...



151121725715, 357044346 (enter hexadecimal values and then press on OCT)



Similarly decimal is ...



14114335693, 4010949011(enter hexadecimal values and then press on DEC)



So required skill-set is "use available resources"



With best regards!

Answer 5

230 ≡ 0 (mod 2) 115 ≡ 1 (mod 2) 57 ≡ 1 (mod 2) 28 ≡ 0 (mod 2) 14 ≡ 0 (mod 2) 7 ≡ 1 (mod 2) 3 ≡ 1 (mod 2) 1 ≡ 1 (mod 2) 11100110 1*2^7 + 1*2^6 + 1*2^5 + 1*2^2 + 1^2^1 = 128 + 64 + 32 + 4 + 2 = 230

Answer 6

34947ABCD.EF123993 should be a base 16 number, just that it shouldn't have a decimal point ( i think ) and it is one of the bases. Number of any base can be converted into any other bases.



Also, the windows calculator has all of those bases.

Answer 7

3*16^8

+4*16^7

+9*16^6

+4*16^5

+7*16^4

+10*16^3

+11*16^2

+12*16^1

+13*16^0

+14*16^-1

+15*16^-2

+1*16^-3

+2*16^-4

+3*16^-5

+9*16^-6

+9*16^-7

+3*16^-8





Why did you ask this question?



Continue in the same way for the other bases.