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Digit sum

In mathematics, the digit sum of a given integer is the sum of all its digits, (e.g.: the digit sum of 84001 is calculated as 8+4+0+0+1 = 13). Digit sums are most often computed using the decimal representation of the given number, but they may be calculated in any other base; different bases give different digit sums, with the digit sums for binary being on average smaller than those for any other base.

Let S(r,N) be the digit sum for radix r of all non-negative integers less than N. For any 2 ≤ r1 < r2 and for sufficiently large N, S(r1,N) < S(r2,N).

The sum of the decimal digits of the integers 0, 1, 2, ... is given by in the On-Line Encyclopedia of Integer Sequences. use the generating function of this integer sequence (and of the analogous sequence for binary digit sums) to derive several rapidly-converging series with rational and transcendental sums.

The concept of a decimal digit sum is closely related to, but not the same as, the digital root, which is the result of repeatedly applying the digit sum operation until the remaining value is only a single digit. The digital root of any non-zero integer will be a number in the range 1 to 9, whereas the digit sum can take any value. Digit sums and digital roots can be used for quick divisibility tests: a natural number is divisible by 3 or 9 if and only if its digit sum (or digital root) is divisible by 3 or 9, respectively.

Digit sums are also a common ingredient in checksum algorithms and were used in this way to check the arithmetic operations of early computers. Earlier, in an era of hand calculation, suggested using sums of 50 digits taken from mathematical tables of logarithms as a form of random number generation; if one assumes that each digit is random, then by the central limit theorem, these digit sums will have a random distribution closely approximating a Gaussian distribution.

The digit sum of the binary representation of a number is known as its Hamming weight or population count; algorithms for performing this operation have been studied, and it has been included as a built-in operation in some computer architectures and some programming languages. These operations are used in computing applications including cryptography, coding theory, and computer chess.

Harshad numbers are defined in terms of divisibility by their digit sums, and Smith numbers are defined by the equality of their digit sums with the digit sums of their prime factorizations. studies the question of how many integers are mapped to the same value by the sum-of-digits function.

6% of the numbers below 100000 have a digit sum of 23, which is along with 22 the most common digit sum within this limit.

From Yahoo Answers

Question:What's (2a+8) + (5a-5) ? I know how to do it but I'm not sure if you add 2+8 then add the 'a' ... Please help! Thank you so much!

Answers:a is a variable. It's different from the constant (#), though it adds and subtracts the same. You just don't group it with the constants. so.... split them up like this (2a+ 5a) + (8+ (-5))= 7a+ 3

Question:2. Find the indicated sums and differences. x+5/3 - x-3/2. 3. Find the indicated sums and differences. 3/a-5 + a+2/a. 4. Find the indicated sums and differences. 3k/k-2 + 6/2-k. 5. Find the indicated sums and differences. b+1/b+2 - b+3/b+4. Thanks so much.

Answers:Okay: 1. Multiply each fraction by the missing denominator. In other words, for the first one, multiply top and bottom by (x-3). Your common denominator here is (x+5)(x-3). So, you get 3x-9-2x-10 all over (x+5)(x-3). So, combine like terms and you get x-19 / (x+5)(x-3) 2. Similar: Instead, multiply each fraction by the missing constant that makes it a common denominator, this case 6. So, multiply the first one top and bottom by 2 and the second one by 3. You get 2x+10-3x+9 all over 6. You get in the end: -x+19 or 19-x / 6. 3. Similar again: multiply the first fraction top and bottom by a and the second one by a-5. So you get 3a+a^2-3a-10 all over a(a-5). Combining like terms, you get a^2-10 over a^2-5a. Notice here if you factor out an a from top and bottom, look what happens: a(a-5) over a(a-5) so these both cancel and you get 1 as your answer. 4. For this one, multiply the second term first by -1. This will make them identical denominators, and then combine: 3k-6 / k-2. Then simplify by factoring out a 3: 3(k-2)/k-2 or 3. 5. Multiply each fraction, as in the earlier problems, by the missing denominator: b^2+5b+4-b^2-5b-6 all over (b+2)(b+4). You can cancel a lot here and you are left with -2 / (b+2)(b+4)

Question:to its right how many such numbers are there? One of my numbers are chosen at random. What is the probability that it is a 1? I circle the middle digit of all the numbers, and add them up. What is the sum? HARDD, PLEASE HELP than the one to its right*

Answers:9876543 8765432 7654321 6543210 The answer is four (4) not 5 not 3 as BILL illustrated but did not answer. The probability (at random) is 1 in 10 (0-9) (however it would only work in one position of two numbers and the odds of that are worse than the lottery like 10000000^10 to 1) 6 + 5 + 4 + 3 + 2 = 20

Question:I worked this one out but wasn't sure which one was correct... (1/p^2 - 2p + 1) - (1/p^2 + p - 2) Is the answer 3/(p-1)^2(p+2) OR p+1/(p-1)(p+2) *I wasn't sure if the LCD of (p-1)(p-1) and (p-1) was just (p-1) or (p-1)^2 Also, what would be the final answer for [(a+b) / (a-b)] + [(a-b) / (a+b)] - [(b-a) / (a-b)] + [(b-a) / (a+b)] Last one... Is this simplified? 2u^2 + 5uv - 2v^2 / (2u+v)(2u-v) THANKS!

Answers:for the first one, you have to get the fractions down to the same denominator before you can add. the least common multiple is (p-1)^2 since (p-1) is not even a multiple of (p-1)^2. all that said, your first answer is right :) for the second one, the important thing to notice is that (a-b) = -(b-a). so the 2nd term cancels with the 4th and we can add the other two since they have the same denominator. final answer = 2a / (a-b) third one looks good :)