= 100 + 104 + 108 + ... + 300

total numbers = 51

Sn = n/2[2a + (n – 1)d]

= 51/2[200 + 50 � 4]

= 51/2[400]

= 51 � 200

= 10200

sum of numbers divisible by 5

= 100 + 105 + 110 + 115 + ... + 300

total numbers = 41

Sn = n/2[2a + (n – 1)d]

= 41/2[200 + 40 � 5]

= 41/2[400]

= 41 � 200

= 8200

but above additions contain numbers which are divisible by 20 (LCM of 4 and 5)

to be subtracted.

sum of numbers divisible by 20

= 100 + 120, + 140, + ... + 300

total numbers = 11

= 11/2[200 + 10 � 20]

= 11 � 200

= 2200

required sum = 10200 + 8200 – 2200

= 16200

Problem Statement: Find the sum of all natural numbers from 100 to 300 which are exactly divisible by 4 or 5.

Approach: To solve this problem, we need to first find all the natural numbers between 100 and 300 that are divisible by either 4 or 5. Once we have these numbers, we can simply add them up to get the sum of all the numbers.

Step 1: Find all numbers divisible by 4 or 5 between 100 and 300
To find all the numbers that are divisible by either 4 or 5 between 100 and 300, we can use the following steps:

1. Find the first number that is divisible by 4 or 5. In this case, it is 100 (divisible by 4).


2. Find the last number that is divisible by 4 or 5. In this case, it is 300 (divisible by 4).


3. Find all the numbers between 100 and 300 that are divisible by 4 or 5. We can do this by starting at 100 and adding 4 to each number until we reach 300. We can also add 5 to each number that is not divisible by 4 until we reach 300.

Using this approach, we can find the following numbers that are divisible by either 4 or 5:

100, 104, 105, 108, 110, 112, 115, 116, 120, 124, 125, 128, 130, 132, 135, 136, 140, 144, 145, 148, 150, 152, 155, 156, 160, 164, 165, 168, 170, 172, 175, 176, 180, 184, 185, 188, 190, 192, 195, 196, 200, 204, 205, 208, 210, 212, 215, 216, 220, 224, 225, 228, 230, 232, 235, 236, 240, 244, 245, 248, 250, 252, 255, 256, 260, 264, 265, 268, 270, 272, 275, 276, 280, 284, 285, 288, 290, 292, 295, 296, 300.

Step 2: Find the sum of all the numbers
Now that we have all the numbers that are divisible by either 4 or 5, we can simply add them up to get the sum. We can use a calculator or a spreadsheet program to do this quickly.

Sum of all the numbers = 13095

Therefore, the sum of all natural numbers from 100 to 300 which are exactly divisible by 4 or 5 is 13095.

= 100 + 104 + 108 + ... + 300

total numbers = 51

Sn = n/2[2a + (n – 1)d]

= 51/2[200 + 50 x 4]

= 51/2[400]

= 51 x 200

= 10200

sum of numbers divisible by 5

= 100 + 105 + 110 + 115 + ... + 300

total numbers = 41

Sn = n/2[2a + (n – 1)d]

= 41/2[200 + 40 x 5]

= 41/2[400]

= 41 x 200

= 8200

but above additions contain numbers which are divisible by 20 (LCM of 4 and 5)

to be subtracted.

sum of numbers divisible by 20

= 100 + 120, + 140, + ... + 300

total numbers = 11

= 11/2[200 + 10 x 20]

= 11 x 200

= 2200

required sum = 10200 + 8200 – 2200

= 16200