a=1; b=1000!;c= b%(12^a); if(c==0, goto4, goto5);printc, a; a++;if(a<1000, goto2, 0);a=1;b++;if(b<1000, goto2, discard=0;
OUTPUT = n =497 or 12^497
+26367 Find the largest positive integer n such that \(12^n\) divides \(1000!\).
\(\begin{array}{|rcll|} \hline \mathbf{2^k =\ ?} \\ \hline k &=& \lfloor\dfrac{1000}{2}\rfloor + \lfloor\dfrac{1000}{2^2}\rfloor+\lfloor\dfrac{1000}{2^3}\rfloor \\ && +\lfloor\dfrac{1000}{2^4}\rfloor+ \lfloor\dfrac{1000}{2^5}\rfloor+\lfloor\dfrac{1000}{2^6}\rfloor \\ && +\lfloor\dfrac{1000}{2^7}\rfloor+ \lfloor\dfrac{1000}{2^8}\rfloor+\lfloor\dfrac{1000}{2^9}\rfloor \\\\ k &=& 500+ 250 +125 \\ && +\lfloor62.5\rfloor + \lfloor31.25\rfloor + \lfloor15.625\rfloor \\ && +\lfloor7.8125\rfloor + \lfloor3.90625\rfloor + \lfloor1.953125\rfloor \\\\ k &=& 500+ 250 +125 +62 +31 \\ && + 15 +7 + 3 +1 \\ \mathbf{k} &=& \mathbf{994} \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{3^m =\ ?} \\ \hline m &=& \lfloor\dfrac{1000}{3}\rfloor + \lfloor\dfrac{1000}{3^2}\rfloor+\lfloor\dfrac{1000}{3^3}\rfloor \\ && +\lfloor\dfrac{1000}{3^4}\rfloor+ \lfloor\dfrac{1000}{3^5}\rfloor+\lfloor\dfrac{1000}{3^6}\rfloor \\\\ m &=& +\lfloor333.\bar{3}\rfloor + \lfloor111.\bar{1}\rfloor + \lfloor37.\bar{037}\rfloor \\ && +\lfloor12.34\ldots\rfloor + \lfloor4.11\ldots\rfloor + \lfloor1.37\ldots\rfloor \\\\ m &=& 333+ 111 + 37 \\ && +12 +4 + 1 \\ \mathbf{m} &=& \mathbf{498} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline 1000! &=& 2^{994} \times 3^{498}\times \ldots \times 997 \quad | \quad \text{prime factors} \\\\ 1000! &=& \left(2^2\right)^{497} \times 3^{498}\times\ldots \\\\ 1000! &=& \left(2^2\right)^{497} \times 3^{497}3 \times\ldots \\\\ 1000! &=& 4^{497} \times 3^{497}3 \times\ldots \\\\ 1000! &=& (4+3)^{497}3 \times\ldots \\\\ 1000! &=& 12^{\mathbf{497}}3 \times\ldots \\ \hline \end{array}\)
The largest positive integer n such that \(12^n\) divides \(1000!\) is 497
