Maria ordered a certain number of televisions at $R \$ 1994.00$ each. She noticed that in the total amount to pay, there are no 0, 7, 8, or 9. What is the smallest number of televisions she ordered?

## Solutions from List 4

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If Maria bought $n$ televisions, then she spent 1994n, which is a multiple of 1994 where the digits $0, 7, 8$, and 9 do not appear. Let's try to limit the value of $n$. First observe that

$$
1994 n=2000 n-6 n
$$

and also that if

$$
6 n<300
$$

then the number $2000 n-6 n$ has 7 or 8 or 9 in the hundreds digit (try some examples, remember that $2000 n$ ends with 3 zeros and then convince yourself). Thus we must have

$$
6 n \geq 300, \quad \text { that is } n \geq 50
$$

Observe that 50 cannot be because the value would end in 0, $\log n \geq 51$. Given that $1994 \times 51=101694$, we have that $n$ cannot be 51 and therefore $n=51+m$ with $m$ positive. Now, since we need the number not to have 0, $1994 m$ must eliminate the 0 from 101694, so $m \geq 4$, but $m=4$ is not a solution because $1994 \times 55$ ends in 0. If we test $n=56$ we have that

$$
1994 \times 56=111664
$$

is the number we are looking for.