Question 35

Show that the cube of a positive integer of the form $6q+r,q$ is an integer and r=0,1,2,3,4,5 is also of the form 6m+r

Solution:

To show that the cube of a positive integer of the form $$n=6q+r$$ (where $$q$$ is an integer and $$r=0,1,2,3,4,5$$) is also of the form $$6m+r$$ for some integer $$m$$, we will follow these steps:

Step 1: Define the number
Let $$n=6q+r$$, where $$q$$ is an integer and $$r$$ can take values $$0,1,2,3,4,5$$.

Step 2: Cube the number
We need to find $$n^3$$:
$${\mathrm{n3}}^{}=(6q+r{)3}^{}$$

Step 3: Expand using the binomial theorem
Using the binomial expansion, we have:
$${\mathrm{n3}}^{}=(6q{)3}^{}+3(6q{)2}^{}\left(r\right)+3\left(6q\right)\left({\mathrm{r2}}^{}\right)+{\mathrm{r3}}^{}$$
This simplifies to:
$${\mathrm{n3}}^{}=216{\mathrm{q3}}^{}+3\cdot 36{\mathrm{q2}}^{}r+3\cdot 6q{\mathrm{r2}}^{}+{\mathrm{r3}}^{}$$
$${\mathrm{n3}}^{}=216{\mathrm{q3}}^{}+\mathrm{108q2r}+\mathrm{18qr2}+r3$$

Step 4: Group the terms
Now, we can group the terms that are multiples of 6:
$${\mathrm{n3}}^{}=6(36{\mathrm{q3}}^{}+\mathrm{18q2r}+\mathrm{3qr2})+{\mathrm{r3}}^{}$$
Let $$m=\mathrm{36q3}+\mathrm{18q2r}+3\mathrm{qr2}$$. Then we can rewrite $${\mathrm{n3}}^{}$$ as:
$${\mathrm{n3}}^{}=6m+{\mathrm{r3}}^{}$$

Step 5: Analyze $$r^3$$
Now, we need to analyze $${\mathrm{r3}}^{}$$ based on the possible values of $$r$$:
– If $$r=0$$, $${\mathrm{r3}}^{}=0$$
– If $$r=1$$, $${\mathrm{r3}}^{}=1$$
– If $$r=2$$, $${\mathrm{r3}}^{}=8$$ (which can be expressed as $$6+2$$)
– If $$r=3$$, $${\mathrm{r3}}^{}=27$$ (which can be expressed as $$24+3$$)
– If $$r=4$$, $${\mathrm{r3}}^{}=64$$ (which can be expressed as $$60+4$$)
– If $$r=5$$, $${\mathrm{r3}}^{}=125$$ (which can be expressed as $$120+5$$)

Step 6: Conclude
In all cases, $${\mathrm{r3}}^{}$$ can be expressed as $$6k+r$$ for some integer $$k$$. Therefore:
$${\mathrm{n3}}^{}=6m+r$$
This shows that the cube of a positive integer of the form $$6q+r$$ is also of the form $$6m+r$$.