Question

Find the sum of all the numbers less than $1000$ and which are neither divisible by $5$ nor by $2$.

Answer 1

Step1: Calculation of the sum of all the numbers divisible by $2$ up to $1000$ .

The sum of first $n$ natural numbers is given by the formula $I_n=\frac{n(n+1)}{2}$.

Numbers divisible by $2$ up to $1000$ are $2,4,6,...1000$.

Sum of all the numbers divisible by $2$ up to $1000$ is given by:

$$\begin{gathered} 2+4+6+\dots +1000=2(1+2+3+..........+500) \\ \Rightarrow 2+4+6+\dots +1000=2\left(\frac{500·501}{2}\right)\left(byu\sin g\frac{n(n+1)}{2}\right) \\ \therefore 2+4+6+\dots +1000=250500equation\left(1\right) \end{gathered}$$

Step2: Calculation of the sum of all the numbers divisible by $5$ up to $1000$ .

Numbers divisible by $5$ up to $1000$ are $5,10,15,\dots 1000$.

Sum of all the numbers divisible by $5$ up to $1000$ is given by:

$$\begin{gathered} 5+10+15+\dots +1000=5(1+2+3+..........+200) \\ \Rightarrow 5+10+15+\dots +1000=5\left(\frac{200·201}{2}\right)\left(byu\sin g\frac{n(n+1)}{2}\right) \\ \therefore 5+10+15+\dots +1000=100500equation\left(2\right) \end{gathered}$$

Step3: Calculation of the number of terms that are divisible by both $2$ and $5$ up to $1000$ .

Numbers divisible by both $2$ and $5$ will be divisible by $10$.

The numbers up to $1000$ which are divisible by $10$ are $10,20,30,40,\dots 990,1000$.

Clearly, this forms an AP with $a=10,d=10,a_n=1000$.

The $n$th term of an A.P. is given by $a_n=a+(n–1)d$, where $a$is a first term and $d$ is a common difference.

Substitute 1000 for $a_n$, 10 for $a$ and 10 for $d$ in the formula $a_n=a+(n–1)d$.

$$\begin{gathered} \Rightarrow 1000=10+(n–1)10 \\ \Rightarrow 990=(n-1)10(subtracting10frombothsides) \\ \Rightarrow 99=n-1(Dividingbothsidesby10) \\ \Rightarrow 100=n(Adding1tobothsides) \\ \therefore n=100 \end{gathered}$$

Step4: Calculation of the sum of all the numbers divisible by both $2$ and $5$ up to $1000$.

The sum of first $n$ terms of an A.P. series is given by the formula $S_n=\frac{n}{2}\left[2a+(n-1)d\right]$, where $a$ is the first term and $d$ is the common difference.

Substitute 10 for $a$, 100 for $n$ and 10 for $d$ in the formula $S_n=\frac{n}{2}\left[2a+(n-1)d\right]$.

$$\begin{gathered} \Rightarrow S_{100}=\frac{100}{2}\left[2\left(10\right)+(100-1)10\right](Substitutingvaluesofa,d,n) \\ \Rightarrow S_{100}=50\left[20+\left(99\right)10\right] \\ \Rightarrow S_{100}=50\left[20+990\right] \\ \Rightarrow S_{100}=50\left[1010\right] \\ \therefore S_{100}=50500equation\left(3\right) \end{gathered}$$

Step5: Calculation of the sum of all the numbers up to $1000$ .

Sum of all the numbers up to $1000$ is given by:

$$\begin{gathered} 1+2+3+\dots +1000=\frac{1001·1000}{2}\left(byu\sin g\frac{n(n+1)}{2}\right) \\ \Rightarrow 1+2+3+\dots +1000=500500equation\left(4\right) \end{gathered}$$

Step6: Calculation of the sum of all the numbers which are neither divisible by $2$ nor by $5$ up to $1000$.

The sum of all the numbers less than $1000$, which are neither divisible by $5$ nor by $2$ is equal to

Sum of all the numbers up to $1000$ – (Sum of all the numbers divisible by $2$ up to $1000$ + Sum of all the numbers divisible by $5$ up to $1000$ – Sum of all the numbers up to $1000$ which are divisible by both $2$ and $5$).

$\begin{array}{rcl}Sumofno.snotdivisibleby(2,5)& =& 500500–(250500+100500–50500)\\ & =& 500500–300500\\ & =& 200000\end{array}$

Final Answer: The sum of all the numbers less than $1000$, which are neither divisible by $5$ nor by $2$ is $200000$.