
1. Consider 8-digit multiples of 4. Is there more of those which contain the digit 1 or those which do not?

(Ján Mazák)


Solution. Let us compute the number of all 8-digit multiples of 4 first. There are 9 possibilities for the first digit of such a number, 10 possibilities for each of the next 5 digits, and such a number has to end with two digits being one of the: 00,04 , $08,12, \ldots, 96$. All together $u=9 \cdot 10^{5} \cdot 25=22500000$ eight-digit multiples of 4 . Similarly there is $v=8 \cdot 9^{5} \cdot 23=10865016$ eight-digit multiples of 4 , which do not contain the digit 1 .

Conclusion. Since $u>2 v$ there is more of 8-digit multiples of 4 which contain number 1 then those which do not.

Remark. It is not necessary to compute $u$ and $v$ to prove $u>2 v$ :

$$
\frac{u}{v}=\frac{9 \cdot 10^{5} \cdot 25}{8 \cdot 9^{5} \cdot 23}=\frac{9}{8} \cdot\left(\frac{10}{9}\right)^{5} \cdot \frac{25}{23}
$$

Using binomial theorem we get:

$$
\left(\frac{10}{9}\right)^{5}=\left(1+\frac{1}{9}\right)^{5}>1+5 \cdot \frac{1}{9}+10 \cdot \frac{1}{9^{2}}=\frac{136}{81}=\frac{8 \cdot 17}{9^{2}}
$$

thus

$$
\frac{u}{v}=\frac{9}{8} \cdot\left(\frac{10}{9}\right)^{5} \cdot \frac{25}{23}>\frac{9}{8} \cdot \frac{8 \cdot 17}{9^{2}} \cdot \frac{25}{23}=\frac{17 \cdot 25}{9 \cdot 23}=\frac{425}{207}>2 .
$$
