Show that the cube of a positive integer of the form \(6q + r\), where \(q\) is an integer and \(r = 0, 1, 2, 3, 4, 5\), is also of the form \(6m + r\).
Yes. The cube of such an integer is again of the form \(6m + r\).
Step 1: Express the integer.
Any integer can be written as
\(n = 6q + r\),
where \(q\) is some integer and \(r\) is the remainder when dividing by 6.
So \(r\) can only be one of the values 0, 1, 2, 3, 4, or 5.
Step 2: Consider the cube.
We need to check the form of \(n^3\). For this, it is enough to look at the values of \(r^3\) modulo 6.
Step 3: Check each remainder one by one.
• If \(r = 0\):
\(0^3 = 0 \equiv 0 \pmod{6}\).
• If \(r = 1\):
\(1^3 = 1 \equiv 1 \pmod{6}\).
• If \(r = 2\):
\(2^3 = 8\).
Since \(8 = 6 + 2\), we have \(2^3 \equiv 2 \pmod{6}\).
• If \(r = 3\):
\(3^3 = 27\).
Since \(27 = 6 \times 4 + 3\), we have \(3^3 \equiv 3 \pmod{6}\).
• If \(r = 4\):
\(4^3 = 64\).
Since \(64 = 6 \times 10 + 4\), we have \(4^3 \equiv 4 \pmod{6}\).
• If \(r = 5\):
\(5^3 = 125\).
Since \(125 = 6 \times 20 + 5\), we have \(5^3 \equiv 5 \pmod{6}\).
Step 4: Observe the pattern.
In each case, we found that
\(r^3 \equiv r \pmod{6}\).
Step 5: Write the final form.
If \(n = 6q + r\), then
\(n^3 = (6q + r)^3\),
and when divided by 6, its remainder is again \(r\).
This means
\(n^3 = 6m + r\) for some integer \(m\).
Conclusion. The cube of any integer of the form \(6q + r\) is again of the form \(6m + r\).