O dobro de um número dividido por 5 deixa resto 1. Qual é o resto da divisão desse número por 5 ?

The double of a number divided by 5 leaves a remainder of 1. What is the remainder of the division of this number by 5?
Solution 1: The double of the number sought is a multiple of 5 increased by 1. Since multiples of 5 end in 0 or 5, the double ends in 1 or 6. But the double is an even number, so it ends in 6. Thus, the number ends in 3 or 8 and, therefore, when divided by 5, leaves a remainder of 3.

Solution 2: We know that the integer $n$ sought satisfies $2 n=5 m+1$, for some integer $m$. Then the product $5 m=2 n-1$ of 5 by $m$ is odd, which implies that $m$ is odd. Thus, $m=2 k+1$, for some integer $k$ and, therefore,

$$
2 n=5 m+1=5(2 k+1)+1=10 k+6=2(5 k+3)
$$

that is, $n=5 k+3$ leaves a remainder of 3 when divided by 5.