Problem 1335

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: $333355$, $333553$, $335533$, $355333$, $553333$, $335335$, $353353$, $533533$, $533335$.