Here is a list of all of the multiples of 4 to 100. 4, 8, 12 ,16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100.
What is the sum of first 100 multiples of 4?
Answer: the sum of 1st hundred multiples of 4 is 20200.
What is the sum of the first multiples of 4?
The multiples of 4 is given by 4 × n. Thus, the multiples of 4 from 16 to 100 = 16, 20, 24……., 100. = (22 ×116)/2 = 2552/2 = 1276. Thus, the sum would be: 16 + 20 + …
What are the common multiples of 12 and 15?
Common Multiples of 12: 12,24,36,48,60,72,…. Common Multiples of 15: 15, 30, 45, 60, 75,… Hence, the Least common multiple of 12 and 15 is 60.
How to calculate the sum of multiples of a number?
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: // This code is contributed by Anant Agarwal. # This code is contributed by Abhishek Agrawal. // This code is contributed by vt_m. ” up to ” . $N . ” = ” .
What is the sum of all the multiples of 3 and 5 below 1000?
We can find x = 166833 and y = 99500. But in x and y both series we have taken the numbers which are both factors of 3 and 5 like 15 so we define Sum of z series is 33165. Now desired numbers is x+ y- z. Final answer is 233168. z is subtr… Loading… Originally Answered: What is the sum of all the multiples of 3 and 5 below 1000?
What is the sum of two numbers less than ten?
If their sum is less than ten then DigitSum (a+b) = DigitSum (DigitSum (a)+DigitSum (b)). But, if the sum of a and b is ten or more then the decimal representation of their sum is a one in the ten’s place and (a+b−10) in the unit’s place.
What are the digits of multiples of nine?
It is well known that the digits of multiples of nine sum to nine; i.e., 9→9, 18→1+8=9, 27→2+7=9, . . ., 99→9+9=18→1+8=9, 108→1+0+8=9, etc. Less well known is that the sum of digits of multiples of other numbers have simple patterns although not so simple as the case of nine. These are shown below:
