Problem No. 19. Divisibility

For which values of $n$, is $2n+5$ divisible by 7? And for which values, is it divisible by 30?

Student tip

Check with the reminders of $n$ when dividing by 7. A number is divisible by 30, if it is divisible by 2, 3 and 5.

Solution

We will explore the divisibility by 7 by checking all possibilities of $n$ modulo 7 i.e.  $n=7k+r, r=0, 1, 2, 3, 4, 5, 6$. Namely, if we substitute this in the expression $2n+5$ we will have
$$2n+5=2 \cdot (7k+r)+5=14k+2 r +5 , r=0, 1, 2, 3, 4, 5, 6.$$
The first part of the expression, $14k$, is divisible by 7 and since the rest of the expression, $2r+5$, is divisible by 7 only for $r=1$, we conclude that the whole expression $2n+5$ is divisible by 7, only for $n=7k+1, k=0, 1, 2, …$.

For the other part of the problem, we have that a number is divisible by 30, if it is divisible by 2, 3 and 5. And since $2n+5=2(n+2)+1$ is an even number, it is not divisible by 2, and consequently it is not divisibly by 30 for any $n$.