A $0,1,1,1,4,4,4,6$ digits can form how many 8-digit numbers that are divisible by 4?
Among the 8-digit numbers that can be formed from the given 8 digits, those divisible by 4 are those whose last two digits are:

$04, 16, 40, 44, 60, 64$.

The number of 8-digit numbers ending in 04 is the number of permutations that can be formed from the remaining 6 elements $(1,1,1,4,4,6)$. This number is

$$
P_{6}^{2,3}=\frac{6!}{2!3!}=\frac{4 \cdot 5 \cdot 6}{2}=60
$$

The number of 8-digit numbers ending in 40 is the same.

Similarly, the number of 8-digit numbers ending in 60 is:

$$
P_{6}^{3,3}=\frac{6!}{3!3!}=20
$$

The number of permutations ending in 16 is $P_{6}^{2,3}=60$, but from this number, we must subtract the number of groups starting with 0: $P_{5}^{2,3}$.

Thus, the number of 8-digit numbers ending in 16 is: $60-\frac{5!}{2!3!}=60-10=50$.

Clearly, the number of 8-digit numbers ending in 64 is the same.

Finally, the number of 8-digit numbers ending in 44 is $P_{6}^{3}$, from which we must subtract the number of groups starting with 0, i.e., $P_{5}^{3}$. Thus, the number of 8-digit numbers ending in 44 is: $\frac{6!}{3!}-\frac{5!}{3!}=120-20=100$.

Therefore, the number of 8-digit numbers that can be formed from the given 8 digits and are divisible by 4 is:

$$
2 \cdot 60+20+2 \cdot 50+100=340
$$

Quittner Pál (Bp. I., Petőfi g. I. o. t.)