In the city of Autolândia, car plate numbers are made up of three digits, ranging from plate 000 to plate 999. To reduce pollution, Mayor Pietro decided to implement a car rotation system, setting the days on which people can use their cars. The rules of the rotation are:

- Monday: only cars with odd-numbered plates;
- Tuesday: only cars with plates whose sum of the three digits is greater than or equal to 11;
- Wednesday: only cars with plates whose number is a multiple of 3;
- Thursday: only cars with plates whose sum of the three digits is less than or equal to 14;
- Friday: only cars with plates containing at least two identical digits;
- Saturday: only cars with a plate number strictly less than 500;
- Sunday: only cars with plates where all three digits are less than or equal to 5.

a) On which days can the car with the plate 729 circulate?

b) Maria, the mayor's wife, wants a car that can circulate every day except Sundays. What plate should she have?

c) Mayor Pietro needs a plate that allows him to circulate every day. What plate should he have?

d) Why can all residents of Autolândia circulate at least once a week?
Solution

a) Since the number 729 is odd, the car with the license plate number 729 can circulate on Mondays. Since $7+2+9=18$, the car can also circulate on Tuesdays, but not on Thursdays. Since 729 is a multiple of 3, this car can also circulate on Wednesdays. Since 729 does not have identical digits, is greater than 500, and has a digit greater than 6, it cannot circulate on Fridays, Saturdays, and Sundays.

b) The reader can verify that the answer to this question is the number 363. We will next show a way to arrive at this answer.

Let $ABC$ be the license plate number of Maria's car. Since the car must circulate on Saturdays, we must have

$$
A \in\{0,1,2,3,4\}
$$

Moreover, the car must circulate on Tuesdays and Thursdays. Therefore,

$$
A+B+C \in\{11,12,13,14\}
$$

Furthermore, to circulate on Mondays, we must have that

$$
C \in\{1,3,5,7,9\}
$$

We will divide the possibilities into four cases:

First case: $A=0$.

In this case, we have that $B+C=A+B+C \in\{11,12,13,14\}$. It is impossible to obtain this equality if $B=0$ or $C=0$. Therefore, we must assume that $B \neq 0$ and that $C \neq 0$. In particular, it holds that $B \neq A$ and $C \neq A$. Since the car must circulate on Fridays, we have that $B=C$. Thus $2C = B+C \in\{11,12,13,14\}$. Since we must have $C \in\{1,3,5,7,9\}$, the only possibility is $ABC=077$. But this number is not a multiple of 3. Thus the car with the license plate number 077 is prohibited from circulating on Wednesdays. Therefore, this case must be discarded.

Second case: $A=1$.

In this case, we have that $B+C=A+B+C-1 \in\{10,11,12,13\}$. Thus, if $B=1$, then we must necessarily have $C=9$. Therefore, we would have $ABC=119$ which is not a multiple of 3, and thus cannot circulate on Wednesdays. On the other hand, if $C=1$, then we must necessarily have $B=9$, obtaining that $ABC=191$ which is also not a multiple of 3. In particular, it holds that $B \neq A$ and $C \neq A$. Since the car must circulate on Fridays, we have that $B=C$. Thus $2C = B+C \in\{10,11,12,13\}$. Since we must have $C \in\{1,3,5,7,9\}$, the only possible choice is $C=5$. Therefore, we have that $ABC=155$ which is not a multiple of 3. Thus, this case must be discarded.

Third case: $A=2$.

In this case, we have that $B+C=A+B+C-2 \in\{9,10,11,12\}$. Thus, if $B=2$, then the only possible choices for $C$ would be $C=7$ or $C=9$. Therefore, we should have $ABC=227$ or $ABC=229$. Since neither of them is a multiple of 3, we must discard the option $B=2$. Also, $C \neq 2$, since $C$ must be odd. In particular, we have that $B \neq A$ and $C \neq A$. Since the car must circulate on Fridays, we have that $B=C$. Thus $2C = B+C \in\{9,10,11,12\}$. Since we must have $C \in\{1,3,5,7,9\}$, the only possible choice is $C=5$. Therefore, we have that $ABC=255$. However, a car with this license plate number would not be prohibited from circulating on Sundays, as it satisfies the requirement that all digits are less than 6. Thus, this case must be discarded.

Fourth case: $A=3$.

In this case, we have that $B+C=A+B+C-3 \in\{8,9,10,11\}$. Thus, if $B=3$, then the only possible choices for $C$ would be $C=5$ or $C=7$. Therefore, we should have $ABC=335$ or $ABC=337$. Since neither of them is a multiple of 3, we must discard the option $B=3$. If $C \neq 3$, then in particular we have that $B \neq A$ and $C \neq A$. For the car to be able to circulate on Fridays, we should have $B=C$. Thus $2C = B+C \in\{8,9,10,11\}$. Since we should have $C \in\{1,3,5,7,9\}$, the only possible choice is $C=5$. Therefore, we would have $ABC=355$, which is not a multiple of 3. Thus, we must also discard the option $C \neq 3$. Since $B+C \in\{8,9,10,11\}$, it holds that