, 2023

The five-digit positive integer $4B5B2$ is divisible by 3 exactly when the
sum of its digits, $4+B+5+B+2=2B+11$
is divisible by 3.

Checking the possible values of $B$,

Value of $\boldsymbol{B}$
1
2
3
4
5
6
7
8
9

Value of $\boldsymbol{2B+11}$
13
15
17
19
21
23
25
27
29

we get that $2B+11$ is divisible
by 3 when $B=2$, $B=5$ and $B=8$.
Solution 1

Since $ABABA$ is not divisible by
3, then $A+B+A+B+A=3A+2B$ is not
divisible by 3.

Since $3A$ is divisible by 3 for all
possible values of the digit $A$,
then if $2B$ were also divisible by 3
(that is, if $B=3$, 6, 9), it would
be the case that $3A+2B$ is divisible
by 3.

Since $3A+2B$ is not divisible by 3,
then $2B$ cannot be divisible by 3
and so the possible values of $B$ are
1, 2, 4, 5, 7, 8.

Since $ABABA$ is divisible by 4, then
the two-digit positive integer $BA$
is divisible by 4.

For each of the possible values of $B$, namely $B=1$, 2, 4, 5, 7, 8, we determine the
values of $A$ for which $BA$ is divisible by 4 .

For example when $B=1$, the two-digit
positive integer $1A$ is divisible by
4 exactly when $A=2$ or $A=6$.

In the table below, we determine the remaining pairs $A$ and $B$ that are possible.

$\boldsymbol{B}$
$\boldsymbol{A}$
$\boldsymbol{(A,B)}$

1
2, 6
$(2,1)$, $(6,1)$

2
4, 8
$(4,2)$, $(8,2)$

4
4, 8
$(4,4)$, $(8,4)$

5
2, 6
$(2,5)$, $(6,5)$

7
2, 6
$(2,7)$, $(6,7)$

8
4, 8
$(4,8)$, $(8,8)$

Thus, there are 12 different pairs of non-zero digits $A$ and $B$ that are possible.

Solution 2

Since $ABABA$ is divisible by 4,
then it is also divisible by 2 and thus even.

Since $ABABA$ is even, then the ones
digit is even, and so the possible values of $A$ are 2, 4, 6, 8.

Since $ABABA$ is divisible by 4, then
the two-digit positive integer $BA$
is divisible by 4.

For each of the possible values of $A$, namely $A=2$, 4, 6, 8, we determine the values
of $B$ for which $BA$ is divisible by 4 .

For example when $A=2$, the two-digit
positive integer $B2$ is divisible by
4 exactly when $B=1, 3, 5, 7, $\text{and}$ }
9.$

When $A=4$, the two-digit positive
integer $B4$ is divisible by 4
exactly when $B=2, 4, 6, $\text{and}$ }
8.$

When $A=6$, the two-digit positive
integer $B6$ is divisible by 4
exactly when $B=1, 3, 5, 7, $\text{and}$ }
9.$

When $A=8$, the two-digit positive
integer $B8$ is divisible by 4
exactly when $B=2, 4, 6, $\text{and}$ }
8.$

Finally, we consider the fact that $ABABA$ is not divisible by 3.

As was shown in Solution 1, the possible values of $B$ are 1, 2, 4, 5, 7, 8 ($B\neq 3, 6, 9$).

Combining this information with the previous values of $A$ and $B$, we get that the possible pairs of
non-zero digits $(A, B)$ are $(2,1)$, $(2,5)$, $(2,7)$, $(4,2)$, $(4,4)$, $(4,8)$, $(6,1)$, $(6,5)$, $(6,7)$, $(8,2)$, $(8,4)$, and $(8,8)$.

Thus, there are 12 different pairs of non-zero digits $A$ and $B$ that are possible.
If $t=ACA2\times BAC$ is
divisible by 15, then $t$ is
divisible by both 5 and 3.

An integer is divisible by 5 exactly when its ones digit is 0 or 5, and
so $ACA2$ is not divisible by
5.

Since $t=ACA2\times BAC$ is divisible
by 5 and $ACA2$ is not, then $BAC$ must be divisible by 5, which means
that $C=5$ (since $C$ is a non-zero digit).

Substituting $C=5$, we get $t=A5A2\times BA5$.

Since $t$ is divisible by 3 and 3
is a prime number, then at least one of $A5A2$ or $BA5$ is divisible by 3, and so $A+5+A+2=2A+7$ is divisible by 3, or $B+A+5$ is divisible by 3, or both are
divisible by 3.

Since $t$ is not divisible by 12,
but $t$ is divisible by 3, then $t$ is not divisible by 4.

The three-digit integer $BA5$ is not
divisible by 2 (and thus not divisible by 4) for all possible values of
$A$, and so the four-digit integer
$A5A2$ must not be divisible by 4
.

The four-digit integer $A5A2$ is
divisible by 4 when the two-digit integer $A2$ is divisible by 4, or when $A=1$, 3, 5, 7, or 9, and so the possible
values of $A$ are 2, 4, 6, 8.

Finally, we return to the requirement that $t=A5A2\times BA5$ is divisible by 3,
meaning that at least one of $2A+7$
or $B+A+5$ is divisible by 3.

When $A=2$, $2A+7=11$ which is not divisible by 3 and
so $B+A+5=B+7$ must be divisible by
3. $B+7$ is divisible by 3 exactly
when $B=2,5$, or 8, and so there are
3 triples $A$, $B$, $C$
that are possible in this case.

When $A=4$, $2A+7=15$ which is divisible by 3, which
means that $B$ can be equal to any
non-zero digit, and so there are 9 triples $A$, $B$, $C$
that are possible in this case.

When $A=6$, $2A+7=19$ which is not divisible by 3 and
so $B+A+5=B+11$ must be divisible by
3. $B+11$ is divisible by 3 exactly
when $B=1$, 4, or 7, and so there are
3 triples $A$, $B$, $C$
that are possible in this case.

When $A=8$, $2A+7=23$ which is not divisible by 3 and
so $B+A+5=B+13$ must be divisible by
3. $B+13$ is divisible by 3 exactly
when $B=2$, 5 or 8, and so there are
3 triples $A$, $B$, $C$
that are possible in this case.

Therefore, there are $3+9+3+3=18$
different triples of non-zero digits $A$, $B$, $C$
that are possible.