Question

the sum of integers from 1 to 100 which are divided by 5 or 3 is​

Answer 1

Answer:

2632

Step-by-step explanation:

sum of all number between 1 to 100 isS=n[2a+(n−1)d]/2 =100[2+(100−1)]/2 =100*101/2 =5050sum of all numbers 3 to 99 multiple of 3s=n[2a+(n−1)d]/2 =33(6+96)/2 =1683Sum of all number5 5 to 100 multiple of 5S=n[2a+(n−1)d]/2 =20(10+95)/2 =1050Sum of all numbers 15 to 90 multiple of 15S=n[2a+(n−1)d]/2 =6[30+75]/2 =315sum of all integer from 1 to 100 not divisible by 3 and 5 is=5050−1683−1050+315 =2632

Answer 2

Answer:

there are 20 numbers divisible by 5 and 33 numbers divisible by 3 so the total numbers which are divisible by both 3 and 5 between 1 to 100 are 20+33=53 numbers.

Step-by-step explanation:

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