Answer: 9
Solution: From the divisibility rule for 11 , we know that the difference of the sum of the odd-positioned digits and the even-positioned digits must be equal to a multiple of 11 . The only way this can happen here is that if 3 s and 5 s are equally distributed over odd and even positions, i.e., two 3 s and one 5 will have odd positions, and two 3 s and one 5 will have even positions. In the odd positions, there are exactly three ways to arrange them, and same for the even positions. Hence, the number of eligible six-digit numbers is $3^{2}=9$. As a check, we can enumerate them all: $333355,333553,335533,355333$, $553333,335335,353353,533533,533335$.
13. How many ways are there to arrange four 3 s and two 5 s into a six-digit number divisible by 11 ?