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.
To find the product of a 2-digit number and a 4-digit number that is approximately 500,000, we need to consider the magnitude of the numbers involved. Since a 2-digit number ranges from 10 to 99 and a 4-digit number ranges from 1000 to 9999, their product will be in the range of 10,000 to 99,000,000. To get a product around 500,000, we can estimate that the 2-digit number is around 50 and the 4-digit number is around 10,000. Therefore, the product of a 50 and 10,000 is 500,000.
The largest two-digit even number is 98. In the decimal system, each digit can range from 0 to 9, but for a two-digit number, the first digit cannot be zero. Since we are looking for an even number, the last digit must be an even number, which leaves us with 0, 2, 4, 6, or 8. Among these options, 8 is the largest even digit, resulting in 98 as the largest two-digit even number.
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
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).
Oh, dude, let me break it down for you. So, to find the number of 5-digit combinations from 1 to 60, you just do 60 minus 1 plus 1, which gives you 60. So, there are like 60 different 5-digit number combinations you can make from that range. Easy peasy, lemon squeezy!
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,.
A possible number range of 900,001 to 900,999 in every case the answer to your question is Zero.