find the sum of all natural numbers less than 1000 and which are neither divisible by 5 nor 3
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We need to find the sum of all the numbers less than 1000, which are neither divisible by 5 nor by 2.
Numbers divisible by 2 upto 1000 are 2, 4 , 6, ........ 1000.
Sum of all the numbers divisible by 2 upto 1000 = 2 + 4 + 6 + ....... + 1000 = 2 (1 + 2 + 3 + .......... + 500)
[Using: sum of first n natural numbers
Numbers divisible by 5 upto 1000 are 5, 10 , 15, ........ 1000.
Sum of all the numbers divisible by 5 upto 1000 = 5 + 10 + 15 + ....... + 1000 = 5 (1 + 2 + 3 + ........ + 200)
Let us find out the sum of all the numbers which are divisible by both 5 and 2.
Numbers divisible by both 2 and 5 will be divisible by 10.
The numbers upto 1000 which are divisible by 10 are: 10, 20, 30, 40, ............ 990, 1000.
Clearly, this forms an AP with a = 10, d = 10, an = 1000, where n can be found out as follows:
an = a + (n – 1) d
⇒ 1000 = 10 + (n – 1) × 10
⇒n = 100

Sum of all the numbers upto 1000 = 1 + 2 + 3 + ........... + 999 + 1000 
Sum of all the numbers less than 1000, which are neither divisible by 5 nor by 2 =
Sum of all the numbers upto 1000 – (Sum of all the numbers divisible by 2 upto 1000 + Sum of all the numbers divisible by 5 upto 1000 – Sum of all the numbers which are divisible by both 2 and 5)
= 500500 – (250500 + 100500 – 50500)
= 200000
Numbers divisible by 2 upto 1000 are 2, 4 , 6, ........ 1000.
Sum of all the numbers divisible by 2 upto 1000 = 2 + 4 + 6 + ....... + 1000 = 2 (1 + 2 + 3 + .......... + 500)
[Using: sum of first n natural numbers
Numbers divisible by 5 upto 1000 are 5, 10 , 15, ........ 1000.
Sum of all the numbers divisible by 5 upto 1000 = 5 + 10 + 15 + ....... + 1000 = 5 (1 + 2 + 3 + ........ + 200)
Let us find out the sum of all the numbers which are divisible by both 5 and 2.
Numbers divisible by both 2 and 5 will be divisible by 10.
The numbers upto 1000 which are divisible by 10 are: 10, 20, 30, 40, ............ 990, 1000.
Clearly, this forms an AP with a = 10, d = 10, an = 1000, where n can be found out as follows:
an = a + (n – 1) d
⇒ 1000 = 10 + (n – 1) × 10
⇒n = 100

Sum of all the numbers upto 1000 = 1 + 2 + 3 + ........... + 999 + 1000 
Sum of all the numbers less than 1000, which are neither divisible by 5 nor by 2 =
Sum of all the numbers upto 1000 – (Sum of all the numbers divisible by 2 upto 1000 + Sum of all the numbers divisible by 5 upto 1000 – Sum of all the numbers which are divisible by both 2 and 5)
= 500500 – (250500 + 100500 – 50500)
= 200000
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