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 \(n\).
$$2n+5=2 \cdot (7k+r)+5=14k+2 r +5 , r=0, 1, 2, 3, 4, 5, 6.$$
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 \(n\).
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