187k views
2 votes
Find the sum of all positive integers less than 100, which are not divisible by 3.

Answer, please

2 Answers

0 votes

Answer:

867

Explanation:

Use the "Gauss" method where you list the addends in reverse order:

Sum = 3+9+15+...+93+99

Sum = 99+93+...+9+3

2*Sum = 102+102+... = 102*17 (102 occurs 17 times)

Sum = 51*17 = 867

User Feuyeux
by
4.8k points
1 vote

Answer: 3,267

Explanation:

The first thing is to find the sum of all positive integers less than 100 and also find the sum of all positive integers that are divisible by three , the difference between the two will give the sum of all positive integers less than 100, which are not divisible by 3.

Sum of all positive integers less than 100 implies

1 + 2 + 3 + 4 + ... + 99 , this means there are 99 numbers altogether .

Using the formula for calculating the sum of terms in Arithmetic series , we have :

Sum = n/2 ( a + l)

where n is the number of terms

a is the first term , and

l is the last term

substituting , we have

Sum = 99/ 2 ( 1 + 99)

Sum = 99/2 (100)

Therefore : the sum of all positive integers less than 100 is 4950

Also , sum of all positive integers divisible by 3 implies :

3 + 6 + 9 +... + 99

There are 33 numbers in all

Also using the formula for calculating the sum of terms in Arithmetic series , we have

Sum = n/2 ( a + l )

Sum = 33/2 ( 3 + 99)

sum = 33/2 ( 102)

Sum = 1683

Therefore :the sum of all positive integers less than 100, which are not divisible by 3 implies

4950 - 1683 = 3,267

User Chris Sparrow
by
4.9k points