Question

How many number lie between $10$ and $300$, which when divided by $4$ leave a remainder $3$?

Answer 1

The first number greater $10$ which when divided by $4$ leaves a remainder $3$ is $11$.
So, the next number will be $11+4=15$
The other numbers in this will be $=15+4=19;19+4=23;23+4=27$
So, the AP$=11,15,19,23,27,\mathrm{32......}$
The last term of this AP will be $299$.
We have to find the total numbers that lie between $10$ and $300$, which when divided by $4$ leaves a remainder $3$.
In fact we have to find '$n$' that is the number of terms of the AP.
$a_n=a+(n-1)d$
Where $a_n=299$, first term $a=11$ and common difference is $d=15-11=4$
Thus, $299=11+(n-1)4$
$299=11+4n-4$
$299-11+4=4n$
$292=4n$
$n=\frac{292}{4}$
$\therefore n=73$
So, total $73$ numbers lie between $10$ to $300$ which when divided by $4$ leaves a remainder $3$.