, 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 $57\,975$.