$$x(x+1)(x+2)\cdots(x+23)=\sum_{n=1}^{24}a_nx^n$$. The number of coefficients $$a_n$$ that are multiples of 3 is

Among the integers $$0,1,\ldots,23$$ there are eight numbers in each residue class modulo $$3$$. The coefficients of $$x^{24},x^{22},\ldots,x^8$$ are not divisible by $$3$$, while the coefficients of $$x^{23},x^{21},\ldots,x^7$$ and all lower powers are divisible by $$3$$. Therefore there are $$24-9=15$$ coefficients that are multiples of $$3$$.