Question

The sum of all natural numbers from 100 to 300 which are exactly divisible by 4 or 5

Answer 1

"Ans: 16200

Assuming that the boundary includes 100 and 300.

Terms divisible by 4 are

100,104,108,⋯,300100,104,108,⋯,300

Number of such terms =300−1004+1=51=300−1004+1=51

Sum of terms divisible by 4

=100+104+⋯+300=512(100+300)=10200  ⋯(1)=100+104+⋯+300=512(100+300)=10200  ⋯(1)

Terms divisible by 5 are

100,105,110,⋯,300100,105,110,⋯,300

Number of such terms =300−1005+1=41=300−1005+1=41

Sum of terms divisible by 5

=100+105+110+⋯+300=412(100+300)=8200  ⋯(2)=100+105+110+⋯+300=412(100+300)=8200  ⋯(2)

LCM(4,5)=20

Number of terms divisible by 20 are

100,120,⋯,300100,120,⋯,300

These terms will be common in (1)(1) and (2)(2)

Number of such terms =300−10020+1=11=300−10020+1=11

Sum of terms divisible by 20

=100+120+140+⋯+300=112(100+300)=2200  ⋯(3)=100+120+140+⋯+300=112(100+300)=2200  ⋯(3)

From (1),(2),(3)

Required sum

=10200+8200−2200=16200

"

Answer 2

Answer:

16200

Step-by-step explanation:

i hope it will help u