【Analysis】First, according to the problem, find the relationship between $A$, $B$, $C$, and $D$, and combine the condition that "the difference between the most and the least number of pencils is less than 5" for analysis to solve the problem.

【Solution】Solution: Let the number of pencils received by $A, B, C, D$ be $A, B, C, D$ respectively. From $B+C=2D$, we know that $C, D, B$ form an arithmetic sequence, with a common difference of $K$; from $A+B=2C$, we know that $A, C, B$ form an arithmetic sequence, with a common difference of $2K$; since the number of pencils among the four people does not differ by more than 4, $K=0$ or 1; if $K=0$, then $4 \times B=53$, but 53 is not a multiple of 4; if $K=1, A>C>D>B$, then $4 \times C-1=53$, but 54 is not a multiple of 4. In summary, $B$ receives 15 pencils. Therefore, the answer is: 15.
8. (10 points) The teacher bought a total of 53 pencils and distributed them among four students, $A$, $B$, $C$, and $D$. The difference between the maximum and minimum number of pencils received is less than 5. If $B$ gives all the pencils to $A$, then $A$ will have twice as many pencils as $C$. If $B$ gives all the pencils to $C$, then $C$ will have twice as many pencils as $D$. From this, we can determine that $B$ received $\qquad$ pencils.