Question

Find the sum of all natural numbers from 1to/1000 which are neither divisible by 2nor by5

Answer 1

Step-by-step explanation:

Sum of all natural numbers 1 to 1000:

$=\frac{n(n+1)}{2}\\=\frac{1000*1001}{2}\\=500500$ -------(i)

Even numbers from 2 to 1000 form an A.P. with 1st term(a) = 2 and common difference(d) = 2.

Total no of even term(n) between 2 and 1000 is 500

Hence sum of the even numbers:

$=\frac{n}{2}[2a+(n-1)d]\\ =\frac{500}{2}[2*2+499*2]\\ =250*1002\\=250500$ ----------(ii)

Similarly, numbers divisible by 5 in between 1 to 1000 also forms an A.P., inn which 1st term(a) = 5 and common difference (d) = 5 and total number of terms(n) = 200

Therefore,

Sum of the numbers that are divisible by 5:

$=\frac{n}{2}[2a+(n-1)d]\\ =\frac{200}{2}[5*2+199*5]\\ =250*1005\\=100500$ ---------(iii)

Now,

In between 1 to 1000 there are some numbers that are even and also divisible by 5, hence they are divisible by 10, and forms an A.P. with first term(a) =10, common difference(d) = 10 and number of terms(n) = 100

∴ Sum of the numbers that are divisible by 10:

$=\frac{n}{2}[2a+(n-1)d]\\ =\frac{100}{2}[10*2+99*10]\\ =50*1010\\=50500$ --------(iv)

So,

Using equations (i),(ii),(iii), and (iv), we can find the sum of the numbers between 1 to 1000 which are not divisible by 2 or 5 = 500500 - (250500 + 100500 - 50500) = 200000

∴Required Sum is 200000

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