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Octal numbers are in the range 0 to 7. Since 111 binary is 7 decimal, every three bits in a binary number can be directly converted to a single octal digit. Thus the 9-bit binary number 101011100 can be split into three groups of three bits, 101 011 100, each of which can be converted to octal, 5 3 4, making the octal representation 5348. If a binary number is not an exact multiple of 3 bits, pad with zeroes until it is. Note that all bases that are a power of 2 are directly related to binary. A single base-4 digit represents two binary digits, while a base-8 digit represents three bits, base-16 every four bits, and so on.
Yes.
BS 381C is not a colour, is a range of 'industrial' colours identified by a second three digit number
I am less than 100 so the range is 01 - 99, but as I am divisible by 2 then I am even. As my tens digit and ones digit are the same then I am a 2 digit number so the range is now 10 - 98. The sum of my digits is 8, my tens digit and my ones digit are the same . . so the only solution is 44.
-128 to 127
Octal numbers are in the range 0 to 7. Since 111 binary is 7 decimal, every three bits in a binary number can be directly converted to a single octal digit. Thus the 9-bit binary number 101011100 can be split into three groups of three bits, 101 011 100, each of which can be converted to octal, 5 3 4, making the octal representation 5348. If a binary number is not an exact multiple of 3 bits, pad with zeroes until it is. Note that all bases that are a power of 2 are directly related to binary. A single base-4 digit represents two binary digits, while a base-8 digit represents three bits, base-16 every four bits, and so on.
Repeatedly divide the number by 8 until the number is zero. Take the remainders from each division (the remainders will always be in the range 0 to 7 inclusive). The first division finds the lowest-order digit, the last finds the highest-order digit. Example: Decimal value: 421 421 / 8 = 52 r 5 52 / 8 = 6 r 4 6 / 8 = 0 r 6 The remainders are 6, 4 and 5, so 421 decimal is 645 octal. To convert from octal to decimal, multiply each octal digit by 8^n, where n is the zero-based order of the digit (0 being the lowest order), then sum the products. Example: Octal number: 645 5 * (8^0) = 5 * 1 = 5 4 * (8^1) = 4 * 8 = 32 6 * (8^2) = 6 * 64 = 384 384 + 32 + 5 = 421 Note that n^0 = 1 for all n>=0.
Yes.
BS 381C is not a colour, is a range of 'industrial' colours identified by a second three digit number
I am less than 100 so the range is 01 - 99, but as I am divisible by 2 then I am even. As my tens digit and ones digit are the same then I am a 2 digit number so the range is now 10 - 98. The sum of my digits is 8, my tens digit and my ones digit are the same . . so the only solution is 44.
30.The first digit can be one of three digits {3, 6, 9} corresponding to the last digit being {1, 2, 3}, and for each of those three digits, the middle digit can be one of ten digits {0 - 9}, making 3 x 10 = 30 such numbers.It is assumed that a 3 digit number is a number in the range 100-999, excluding numbers starting with a leading zero, eg 090 is not considered a 3 digit number (though it would be a valid 3-digit number for a combination lock with 3 digits).
For any integer greater than 5, where the units digit is 5 then that number can be expressed as the product of n and 5. As such, the number is composite. Therefore all numbers in the range 150 to 200 that have a units digit of 5 are composite,.
-128 to 127
Binary is a base 2 number system, while octal is base 8. This happens to make conversion between binary and octal fairly trivial, although more complex than conversion to hexadecimal. To convert to octal from binary, take each three bits, starting from the least significant bit, and convert them to their octal equivalent. Examples: 25510 = 111111112 = 11 111 111 = 3778 17410 = 101011102 = 10 101 110 = 2568 You can repeat this process for as many bits as you need. A 24-bit number should translate into 8 octal numbers, for reference.
A possible number range of 900,001 to 900,999 in every case the answer to your question is Zero.
I can't say for certain what your specific difficulty is with the process, so I will guess unfamiliarity. There are many fine websites that will perform those calculations automatically.------------------------------There is a general method to convert from base 10 to any other base:divide the number by the base to get a whole number quotient and remaindernote the remainderreplace the number by the quotientif the number is not zero repeat from step 1write the remainders in reverse order to get the decimal number in the new base.With this converting a decimal number to binary is quite straight forward; for example 205 in binary:205 ÷ 2 = 102 r 1102 ÷ 2 = 51 r 051 ÷ 2 = 25 r 125 ÷ 2 = 12 r 112 ÷ 2 = 6 r 06 ÷ 2 = 3 r 03 ÷ 2 = 1 r 11 ÷ 2 = 0 r 1→ 205 in decimal is 1100 1101 in binary.What you may be complaining about is that converting octal and hexadecimal numbers to binary is extremely straight forward and direct; examples:0315 (octal) = 11 001 101 = 1100 1101 in binary0xcd (hexadecimal) = 1100 1101 binaryThese conversions are extremely easy as each digit of an octal or hexadecimal number uses an exact number of binary digits:octal numbers 0-7 are the fill range of the binary numbers 000-111 - 3 binary digitshexadecimal numbers 0-f are the full range of the binary numbers 0000-1111 - 4 binary digits.There is no waste so each digit of an octal or hexadecimal number can be converted into binary directly. Each new octal or hexadecimal place value column is represented by an exact 3 or 4 block of binary digits, so when a place value is added, another block of binary digits is added, so 07 + 01 = 010 which in binary is 111 + 001 = 001 000; similarly 0xf + 0x1 = 0x10 which in binary is 1111 + 0001 = 0001 0000With decimal numbers, however, the digits 0-9 are represented by the binary 0000-1001; if each digit of a decimal number was converted to binary (an encoding known as Binary Coded Decimal, or BCD) then the binary numbers 1010-1111 (6 of them) are not being used and wasted. Alternatively, when a new place value is needed in decimal the binary will still likely use the binary digits already being used without the need for an extra block, eg 9 + 1 = 10 which in binary is 1001 + 0001 = 1010; there is no 1:1 correspondence between blocks of binary digits and decimal digits that occurs with octal and hexadecimal numbers.
There are 544,320 different possibilities. They range all the way from 1,023,456 to 9,876,543 .