Sum of multiples of a number up to N
Given a number a and limit N. Find the sum of multiple of a upto N.
Input : a = 4, N = 23 Output : sum = 60 [Multiples : 4, 8, 12, 16, 20] Input :a = 7, N = 49 Output :sum = 196 [Multiples: 7, 14, 21, 28, 35, 42, 49]
The basic idea is to iterate from i = a to i = n, i++ and check whether i % a == 0 or not.If zero then add i to sum(initially sum = 0).Thus we will get the sum.It will take O(n) time.
We can modify the loop as i = a, i <= n, i = i + a to reduce the number of iterations.But it will also take O(m) time if there is m multiples of a.
To get the result in O(1) time we can use the formula of summation of n natural numbers.For the above example,
a = 4 and N = 23, number of multiples of a, m = N/a(integer division). The multiples are 4, 8, 12, 16, 20.
We can write it as 4 X [1, 2, 3, 4, 5]. So we can get the sum of multiples as:
sum = a * (Summation of 1 to m [natural numbers from 1 to m]) sum = 4 * (m*(m+1) / 2) sum = 4 * (5*6 / 2) = 4 * 15 = 60
Sum of multiples of 7 upto 49 = 196
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