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The sum of digits function is closely related to the theory of partitions and compositions. A partition of n is a sum d1 + d2 + ... + dk, for some k less than or equal to n such that the sum is n, and the d's are ordered from largest to smallest. For example, the partitions of 4 are 4,3+1,2+2,2+1+1 and 1+1+1+1.

Step 1: Take a long number as input from the user. Step 2: Extract a single digit from a number. Step 3: Check if digit is odd or not. Step 4: If digit is odd, then add the digit to variable sum.

An alternating sum is a series of real numbers in which the terms alternate sign. Alternating sums also arise in other cases. For instance, the divisibility rule for 11 is to take the alternating sum of the digits of the integer in question and check if the result is divisble by 11.

Algorithm ? Take a string input from user and store it in the variable called ?s?. Take a i for loop start from i=0 to i<s. length(). Check the condition that character is digit or not. Calculate sum of each digit and store it in the sum variable. After that simply print the value of sum.

Let's solve it together: We want to iterate over the digits, so let's iterate over a string and not a long. ... You should start iterate from the end. ... We want to add up only odd digits, so our update statement should decrement i in two every step: for (int i = lastIndex; 0 <= i; i -= 2)

ALGORITHM: STEP 1: START. STEP 2: INITIALIZE arr[] = {1, 2, 3, 4, 5} STEP 3: SET sum = 0. STEP 4: length= sizeof(arr)/sizeof(arr[0]) STEP 5: SET i=0. REPEAT STEP 6 and STEP 7 UNTIL i<length. STEP 6: sum = sum + arr[i] STEP 7: i=i+1. STEP 8: PRINT "Sum of all the elements of an array:" by assigning sum.

We can calculate the sum of digits of a number by adding a number's digits while ignoring the place values. So, if we have the number 567, we can calculate the digit sum as 5 + 6 + 7, which equals 18.