, 2026

Since $0 < A < B < C < 10$, the largest possible value of $A$ is $7$, and when $A=7$, we must have $B=8$ and $C=9$. Therefore the largest possible peak number is $78\, 987$. Similarly, the smallest possible peak number is $12\, 321$. Therefore the positive difference is $78\,
987-12\,321 = 66\, 666$. Since$36\,245 < ABCBA <
45\,932$, we have$A=4$or$A=3$. If$A=4, then the first inequality is true for all possible values of

B$and$C$, so we only need to consider$4BCB4 < 45\,932$. We must then have$B=5$, and then$C$can be any of$6$,$7$, or$8$, so there are$3$peak numbers when$A=4$. (They are$45\, 654$,$45 \, 754$,$45 \, 854$.) If$A=3, then the second inequality is true for all possible values of

B$and$C$, so we only need to consider$36\,245 < 3BCB3$. Then$B$can be$6$, in which case$7 $\leq C \leq$ 9$, or$B = 7$, in which case$8 $\leq C \leq$ 9$, or$B=8, in which case we have that

C=9$. Altogether, there are$6$peak numbers when$A=3$. (They are$36\,763$,$36\,863$,$36\,963$,$37\,873$,$37\,973$,$38\,983.) Putting both cases together, there are a total of

3+6 = 9 peak numbers satisfying the desired inequalities. A number is a multiple of

15$exactly when it is a multiple of both$5$and$3$, and a number is a multiple of$5 exactly when it has units digit

0$or$5$. Since$0<A<10$and$ABCBA$is a multiple of$15$, we must have$A=5$. Furthermore,$5BCB5$is a multiple of$15$exactly when$5BCB5$is a multiple of$3$. A positive integer is a multiple of$3 exactly when the sum of its digits is a multiple of

3$. That is,$5BCB5$is a multiple of$3$exactly when$10 + 2B + C$is a multiple of$3$. Since$A=5$and$A<B<C$, the possible values of$B$are$6$,$7$and$8$. When$B=6$,$10+12+C$is a multiple of$3$precisely when$C=8 (among the possible values $7$, $8$, $9$ of $C$). When

B=7$,$10+14+C$is a multiple of$3$precisely when$C=9$. Finally, when$B=8, the only possible value of

C$is 9, and$10+16+9$is not a multiple of$3. Therefore, the peak numbers that are a multiple of

15$are$56\,865$and$57\,975$.