Question

The sum of integers from 1 to 100 which is not divisible by 3 or 5 is
$
  a.{\text{ }}2849 \\
  b.{\text{ }}4375 \\
  c.{\text{ }}2317 \\
  d.{\text{ }}2632 \\
$

Answer 1

Hint: - Sum of numbers which is not divisible by 3 or 5 is = sum of all numbers - sum of numbers which is divisible by 3 - sum of numbers which is divisible by 5 + sum of numbers which is divisible by both 3 and 5
The set of numbers which is divisible by 3 from 1 to 100 is $\left\{ {3,6,9,.................,99} \right\}$
As you see this series form an A.P with its first term$\left( {{a_1} = 3} \right)$, last term $\left( {{a_l} = 99} \right)$and common difference $\left( {d = 6 - 3 = 3} \right)$
Therefore number of terms in this series is
${a_l} = {a_1} + \left( {n - 1} \right)d$Where n is the number of terms.
$
   \Rightarrow 99 = 3 + \left( {n - 1} \right)3 \\
   \Rightarrow n - 1 = 32 \\
   \Rightarrow n = 33 \\
$
So, the sum of this series is
$
  {S_n} = \frac{n}{2}\left( {{a_1} + {a_l}} \right) \\
   \Rightarrow {S_n} = \frac{{33}}{2}\left( {3 + 99} \right) = \frac{{33}}{2} \times 102 = 1683 \\
$
The set of numbers which is divisible by 5 from 1 to 100 is $\left\{ {5,10,15,.................,100} \right\}$
As you see this series form an A.P with its first term$\left( {{a_1} = 5} \right)$, last term $\left( {{a_l} = 100} \right)$and common difference $\left( {d = 10 - 5 = 5} \right)$
Therefore number of terms in this series is
${a_l} = {a_1} + \left( {n - 1} \right)d$Where n is the number of terms.
$
   \Rightarrow 100 = 5 + \left( {n - 1} \right)5 \\
   \Rightarrow n - 1 = 19 \\
   \Rightarrow n = 20 \\
$
So, the sum of this series is
$
  {S_n} = \frac{n}{2}\left( {{a_1} + {a_l}} \right) \\
   \Rightarrow {S_n} = \frac{{20}}{2}\left( {5 + 100} \right) = 10 \times 105 = 1050 \\
$
The set of numbers which is divisible by both 3 and 5.
Therefore L.C.M of 3 and 5 is 15
The set of numbers which is divisible by 15 from 1 to 100 is $\left\{ {15,30,.................,90} \right\}$
As you see this series form an A.P with its first term$\left( {{a_1} = 15} \right)$, last term $\left( {{a_l} = 90} \right)$and common difference $\left( {d = 30 - 15 = 15} \right)$
Therefore number of terms in this series is
${a_l} = {a_1} + \left( {n - 1} \right)d$Where n is the number of terms.
$
   \Rightarrow 90 = 15 + \left( {n - 1} \right)15 \\
   \Rightarrow n - 1 = 5 \\
   \Rightarrow n = 6 \\
$
So, the sum of this series is
$
  {S_n} = \frac{n}{2}\left( {{a_1} + {a_l}} \right) \\
   \Rightarrow {S_n} = \frac{6}{2}\left( {15 + 90} \right) = 3 \times 105 = 315 \\
$
The sum of numbers from 1 to 100.
Total number of terms from 1 to 100 is 100.
$
  {S_n} = \frac{n}{2}\left( {{a_1} + {a_l}} \right) \\
   \Rightarrow {S_n} = \frac{{100}}{2}\left( {1 + 100} \right) = 50 \times 101 = 5050 \\
$
Therefore Sum of numbers which is not divisible by 3 or 5 is = sum of all numbers – sum of numbers which is divisible by 3 – sum of numbers which is divisible by 5 + sum of numbers which is divisible by both 3 and 5
$ \Rightarrow S = 5050 - 1683 - 1050 + 315 = 2632$
Hence, option (d) is correct.

Note: - In such types of question always remember some of the basic formulas of A.P which is stated above, then calculate the sum of all the series which is divisible by 3 , 5 and both, then using the formula which is stated above we will get the required sum which is divisible by 3 or 5.