Basket $1$ has $10$ peaches. Basket $2$ has 10 apples and $1$ pear. Basket $3$ to $6$ have $10$ in total.

Let $a = apple, b = peach, c = pear$. $$\begin{cases} \sum_{i=1}^{6} a_i + \sum_{i=1}^{6} b_i + \sum_{i=1}^{6} c_i &= 31 \\ \sum_{i=1}^{6} b_i &= 5 \sum_{i=1}^{6} a_i \\ \sum_{i=1}^{6} a_i &= 5 \sum_{i=1}^{6} c_i \end{cases} $$$$\implies \sum_{i=1}^{6} a_i = 5, \sum_{i=1}^{6} b_i =25, \sum_{i=1}^{6} c_i =1 $$One possible arrangement is $$(apple, peaches, pears) = (0, 5, 1), (1, 4, 0), (1, 4, 0), (1, 4, 0), (1, 4, 0), (1, 4, 0)$$

$$\sum_{i=1}^{6} a_i + \sum_{i=1}^{6} b_i + \sum_{i=1}^{6} c_i = 31 \sum_{i=1}^{6} c_i$$Thus, the total number of fruits must be multiple of $31$.