Question

The sum of all natural numbers less than 200 that are divisible neither by 3 nor by 5 is?
(a) 10730
(b) 10732
(c) 15375
(d) None of these

Answer 1

Find the sum of all the natural numbers by using the formula of sum of n terms of an A.P given as ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$. Here, ‘a’ is the first term, d is the common difference of the A.P and ${{S}_{n}}$ denotes the sum of n terms of an A.P. Now, consider the sum of all the natural numbers less than 200 that are divisible by 3 or 5. For this, consider the multiples of 3 and 5 one by one and find the sum of their multiples. Now, subtract the multiples of sum of multiples of 15 from the sum of multiples of 3 and 5. Finally, subtract this obtained difference from ${{S}_{n}}$ to get the answer.

Complete step by step solution:
Here we have been provided with the natural numbers less than 200 and we are asked to find the sum of all those natural numbers that are not divisible by 3 or 5.
Now, we have 199 natural numbers starting from 1 and less than 200. Here, we will subtract the sum of all the natural numbers less than 200 that are divisible by either 3 or 5 from the sum of all these 199 natural numbers.
These 199 natural numbers will form an A.P with 1 as the first term, 1 as the common difference and 199 as the last term, so using the formula of sum of n terms of an A.P given as ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$, where ‘a’ is the first term, d is the common difference of the A.P and ${{S}_{n}}$ denotes the sum of n terms of an A.P, we get,
$\begin{aligned} & \Rightarrow {{S}_{199}}=\dfrac{199}{2}\left[ 2\left( 1 \right)+\left( 199-1 \right)1 \right] \\\ & \Rightarrow {{S}_{199}}=\dfrac{199}{2}\left[ 2+198 \right] \\\ & \Rightarrow {{S}_{199}}=19900..............\left( i \right) \\\ \end{aligned}$
Now, let us determine the sum of all the multiples of natural numbers divisible by 3 or 5. Here, multiples of 3 will form an A.P with the first term as 3, common difference as 3. Let us assume that there are p such multiples of 3, so mathematically we must have ${{p}^{th}}$ multiple of 3 less than 200, so using the formula for ${{p}^{th}}$ term given as ${{T}_{p}}=a+\left( p-1 \right)d$ we get,
$\begin{aligned} & \Rightarrow 3+\left( p-1 \right)3<200 \\\ & \Rightarrow 3p<200 \\\ & \Rightarrow p<66.67 \\\ \end{aligned}$
Therefore, the value of p must be 66, so the sum of all the multiples of 3 can be given by the sum of 66 terms of the A.P. Therefore we have,
$\begin{aligned} & \Rightarrow {{S}_{66}}=\dfrac{66}{2}\left[ 2\left( 3 \right)+\left( 66-1 \right)3 \right] \\\ & \Rightarrow {{S}_{66}}=33\left[ 6+195 \right] \\\ & \Rightarrow {{S}_{66}}=6633..............\left( ii \right) \\\ \end{aligned}$
Similarly, the multiples of 5 will form an A.P with the first term as 5, common difference as 5. Let us assume that there are q such multiples of 5, so mathematically we must have ${{q}^{th}}$ multiple of 5 less than 200, so using the formula for ${{q}^{th}}$ term given as ${{T}_{q}}=a+\left( q-1 \right)d$ we get,
$\begin{aligned} & \Rightarrow 5+\left( q-1 \right)5<200 \\\ & \Rightarrow 5q<200 \\\ & \Rightarrow q<40 \\\ \end{aligned}$
Therefore, the value of q must be 39, so the sum of all the multiples of 5 can be given by the sum of 39 terms of the A.P. Therefore we have,
$\begin{aligned} & \Rightarrow {{S}_{39}}=\dfrac{39}{2}\left[ 2\left( 5 \right)+\left( 39-1 \right)5 \right] \\\ & \Rightarrow {{S}_{39}}=\dfrac{39}{2}\left[ 10+190 \right] \\\ & \Rightarrow {{S}_{39}}=3900..............\left( iii \right) \\\ \end{aligned}$
Now, the multiples of 15 will be common in both the multiples of 3 and 5, so if we add equations (ii) and (iii) we will get multiples of 15 added two times, therefore we need to subtract the sum of multiples of 15 one time. Here, the multiples of 15 will form an A.P with first term as 15, common difference as 15. Let us assume there are r such multiples of 15, so we must have ${{r}^{th}}$ multiple of 15 less than 200, so using the formula for ${{r}^{th}}$ term given as ${{T}_{r}}=a+\left( r-1 \right)d$ we get,
$\begin{aligned} & \Rightarrow 15+\left( r-1 \right)15<200 \\\ & \Rightarrow 15r<200 \\\ & \Rightarrow r<13.33 \\\ \end{aligned}$
Therefore, the value of r must be 13, so the sum of all the multiples of 15 can be given by the sum of 13 terms of the A.P. Therefore we have,
$\begin{aligned} & \Rightarrow {{S}_{13}}=\dfrac{13}{2}\left[ 2\left( 15 \right)+\left( 13-1 \right)15 \right] \\\ & \Rightarrow {{S}_{13}}=\dfrac{13}{2}\left[ 30+180 \right] \\\ & \Rightarrow {{S}_{13}}=1365..............\left( iv \right) \\\ \end{aligned}$
Therefore, the sum of natural numbers less than 200 which are divisible by 3 or 5 will be given by the adding equations (ii) and (iii) and subtraction equation (iv) from it. So we get,
$\Rightarrow$ Sum of multiples of natural numbers less than 200 divisible by 3 or 5 = 6633 + 3900 – 1365 = 9168
Therefore, the sum of natural numbers less than 200 which are neither divisible by 3 nor 5 will be given by the difference of equation (i) and 9165, so we get,
$\therefore$ Sum of multiples of natural numbers less than 200 which are neither divisible by 3 nor 5 = 19900 – 9168 = 10732

So, the correct answer is “Option b”.

Note: Note that you must not include the natural number 200 in the calculations of the number of terms because it is clearly written that we have to consider all the natural numbers less than 200. Do not forget to subtract multiples of 15 because the L.C.M of 3 and 5 is 15 and both will contain the multiples of 15 which in addition will contain twice of these numbers.