What is the sum of all the numbers which are divisible by 3?

What is the sum of all the numbers which are divisible by 3?

The divisibility rule of 3 for large numbers states that if the sum of all digits of a large number is divisible by 3 or is a multiple of 3 then we can say that the large number is also divisible by 3. Here, the sum of all the digits = 2+2+0+0+7+7 = 18.

What is the summation of the first 100 numbers that are divisible by 3 without any remainder?

Input parameters & values: The number series 3, 6, 9, 12, 15, 18, 21, . . . . , 300. Therefore, 15150 is the sum of first 100 positive integers which are divisible by 3.

Is the sum divisible by 3?

A number is divisible by 3 if the sum of its digits is divisible by 3. For large numbers this rule can be applied again to the result. A.) 19,869,854,568: 1+9+8+6+9+8+5+4+5+6+8 = 69, 6 + 9 = 15, 1 + 5 = 6, so it is divisible by 3.

What are the composite numbers from 50 to 100?

50,51,52,54,55,56,57,58,60,62,63,64,65,66,68,69,70,72,74,75,76,77,78,80,81, 82,84,85,86,88,89,90,91,92,93,94,95,96,98 and 99.

What is the sum of the integers from 1 to 100 that are divisible by 2 or 5?

The integers, which are divisible by both 2 and 5, are 10, 20, … 100. This also forms an A.P. with both the first term and common difference equal to 10. Thus, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.

Is a number divisible by 3 if the sum of its digits is divisible by 3?

Answer: Yes, a number is divisible by 3 if the sum of its digits is divisible by 3. Let’s see the solution with the help of some examples. Explanation: Hence, 27 is divisible by 3.

Why is a number divisible by 3 if its digits sum is divisible by 3?

All but the first and last terms cancel. Thus any power of 10 less 1 is divisible by 9, and therefore also by 3. Every term on the right other than the sum of the digits is divisible by 3. So the remainder when dividing the original number by 3 and the sum of the digits by 3 must be the same.

What is the sum of numbers between 50 and 100?

Sum of numbers from 50 to 100 =count* average. Count= 100–50+1=51. Average=(first+last)/2= 75. Sum = 51*75=3825. Numbers between 50 and 100 not divisible by 2 or 3.

How many numbers between 100 and 500 are divisible by 3 or 7?

So this means the number of numbers divisible by 3 is (3 numbers per 10) (10 numbers per 100) (5 100s in 500) There look to be 14 numbers per 100. If this is the case, and since there are 5 100s in 500, this means there are 14 (5) = 70 numbers between 100 and 500 that are divisible by 7.

Is the sum of 2 or 3 in 1 to 100 numbers?

We know that 100 is not a multiple of 3, but 99 is. So we now know that the trio of “1, 2, 3”s will end at the 99th place, and the 100th place must be 1. Great. So, we have 33 groups of “1, 2, 3”, plus a “1” leftover.

How to find the sum of 2 or 5 integers?

Sum of integers divisible by 2 or 5 = Sum of integers divisible by 2 + Sum of integers divisible by 5 – Sum of integers divisible by 2 & 5 Finding sum of numbers from 1 to 100 divisible by 2 Integers divisible by 2 between 1 to 100 are 2, 4, 6, 8, …100 This forms an A.P. as difference of consecutive terms is constant.