, 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.