Question

The sum of all natural numbers from 100 to 300 which are exactly divisible by 4 and 5 is
a) 2,200
(b) 2,000
(c) 2,220
(d) none of these
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Answer 1

Answer:

None of these

Step-by-step explanation:

Sum of numbers divisible by 4 = 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