Can someone check my solution for this question?
$28! = 2^a \times 3^b \times 5^c \times 7^d \times 11^e \times 13^f \times 17^g \times 19^h \times 23^i$
To find a:
$28/2 = 14$
$28/2^2 = 7$
$28/2^3 = 3$
$28/2^4 = 1$
Where $a= 14+7+3+1= 25$ giving $2^{25}$
I repeated this method to find b to i to give a final answer of
$28! = 2^{25} \times 3^{13} \times 5^6 \times 7^4 \times 11^2 \times 13^2 \times 17 \times 19 \times 23$
Is this method/answer correct?
Your answer is correct and your method is fine, but not your choice of notation. It is not true that $\dfrac{28}{2^3}=3$. You are interested in integer division here, and therefore what you should write is $\left\lfloor\dfrac{28}{2^3}\right\rfloor=3$.