Integer of form 6k + 5 is of form 3k + 2 but not Conversely

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Theorem

Let $n \in \Z$ be an integer of the form:

$n = 6 k + 5$

where $k \in \Z$.

Then $n$ can also be expressed in the form:

$n = 3 k + 2$

for some other $k \in \Z$.


However it is not necessarily the case that if $n$ can be expressed in the form:

$n = 3 k + 2$

then it can also be expressed in the form:

$n = 6 k + 5$


Proof

\(\ds n\) \(=\) \(\ds 6 k + 5\)
\(\ds \) \(=\) \(\ds 3 \paren {2 k} + 3 + 2\)
\(\ds \) \(=\) \(\ds 3 \paren {2 k + 1} + 2\) for some $2 k + 1 \in \Z$

Replacing $2 k + 1$ with $k$ gives the result.

$\Box$


Now consider $n = 8$.

We have that:

$8 = 3 \times 2 + 2$

and so can be expressed in the form:

$n = 3 k + 2$

However:

$8 = 6 \times 1 + 2$

and is in the form:

$n = 6 k + 2$

and so cannot be expressed in the form:

$n = 6 k + 5$

$\blacksquare$


Sources