Question

The sum of integers from $1$ to $100$ should be divisible by $2$ or $3$ is:

• A. $1683$
• B. $4233$
• C. $3417$
• D. $816$

Answer 1

The correct option is C $3417$

$formulaused$
$1+2+3+4+...........+n=\frac{n(n+1)}{2}$
$sumof\text{integers that are divisible by 2}between1and100$
$2+4+6+........100$
$S_1=2(1+2+3+.......+50)$
$=\frac{2\left(50\right)\left(51\right)}{2}$
$=2550$
$sum\text{of integers that are divisible by 3}$
$3+6+9+........+99$
$S_2=3(1+2+3+.........+33)$
$=\frac{3\left(33\right)\left(34\right)}{2}$
$=1683$
$S\text{um of integers that are divisible}\text{by both 2 and 3}$
$\text{6+12+18+}...........\text{96}$
${\text{S}}_3=6(1+2+3+.........+16)$
$=\frac{6\left(16\right)\left(17\right)}{2}$
$=816$
$Bysettheory$
$n(A\cap B)=n\left(A\right)+n\left(B\right)-n(A\cup B)$
$=2550+1683-816$
$=3417$