A $0,1,3,5,6$ digits can form how many different four-digit numbers, all digits being distinct, that are divisible by 15?
Numbers divisible by 15 are those divisible by 5 and 3, meaning they end in 0 or 5, and the sum of their digits is divisible by 3. Due to the latter condition, only the following groups of digits meet the criteria: a) $1,3,5,6$, b) $0,1,5,6$, and c) $0,1,3,5$.

In case a), the number of numbers ending in 5 is $3!=6$.

In case b), the number of numbers ending in 0 is $3!=6$, and the number of numbers ending in 5 is $3!-2!=4$, because we must exclude the groups starting with 0.

In case c), similarly to case b), $6+4=10$ numbers can be formed that meet our conditions.

Therefore, a total of $6+10+10=26$ numbers, each with distinct digits, can be formed from the given digits that are divisible by 15.

Beke Éva and Mária (Bp. XIII., Ép. gép. techn. II. o. t)