Prove that one of any three consecutive positive integers must be divisible by 3.
Yes, in every set of three consecutive integers, one of them is divisible by 3.
Step 1: Represent three consecutive integers.
Let the three consecutive positive integers be:
\(n,\; n+1,\; n+2\)
Step 2: Recall the property of division by 3.
When any integer is divided by 3, the possible remainders are:
\(0,\; 1,\; \text{or } 2\)
Step 3: Check each case for \(n\).
Case A: If \(n \equiv 0 \pmod{3}\), then \(n\) itself is divisible by 3.
Case B: If \(n \equiv 1 \pmod{3}\), then
\(n+2 \equiv 0 \pmod{3}\)
So, \(n+2\) is divisible by 3.
Case C: If \(n \equiv 2 \pmod{3}\), then
\(n+1 \equiv 0 \pmod{3}\)
So, \(n+1\) is divisible by 3.
Step 4: Conclude.
In every group of three consecutive integers, one of them must be divisible by 3. This proves the statement.