The sum of all the elements of the set $\{\alpha \in \{1,2,\dots .,100\}:HCF(\alpha ,24)=1\}$ is

Let $n\left(a\right)=$ number of numbers divisible by a, so
$n\left(2\right)=\{2,4,6\dots ..100\}\Rightarrow 50$ numbers
$n\left(3\right)=\{3,6,9\dots ..99\}\Rightarrow 33$ numbers
$n(2\cap 3)=\{6,12,18\dots ..96\}\Rightarrow 16$ numbers
$\Rightarrow$ Sum of all numbers divisible by $2=\frac{50}{2}[2+100]=50\times 51$
Sum of all numbers divisible by $3=\frac{33}{2}[3+99]=33\times 51$
Sum of all numbers divisible by $6=\frac{16}{2}[6+96]=16\times 51$
Sum of all numbers divisible by either 2 or $3=50\times 51+33\times 51-16\times 51=67\times 51$
Sum of all natural numbers form 1 to $100=\frac{100}{2}[1+100]=50\times 101$
Sum of required values of 'a' is $101\times 50-67\times 51=1633$

The sum of all the elements of the set $\{\alpha \in \{1,2,\dots .,100\}:HCF(\alpha ,24)=1\}$ is 1633.