Question

Prove that for every n E N, when we divide n2 by 7, the remainder is either 0, 1, 2, or 4. In other words, the remainder will never be 3, 5, or 6.

Answer 1

Answer:

For every n E N, when we divide n² by 7, the remainder is either 0, 1, 2, or 4.

Step-by-step explanation:

Given information: n∈N and $\frac{n^2}{7}$.

To prove: For every n E N, when we divide n2 by 7, the remainder is either 0, 1, 2, or 4.

Proof:

Using basic remainder remainder theorem,

$remainder(\frac{n^2}{7})=(remainder(\frac{n}{7})\times remainder(\frac{n}{7}))(\text{mod }7)$

where, mod 7 is modulo 7. It means the remainder after dividing by 7.

If a natural number divide by 7 then the possible remainders are 0,1,2,3,4,5 and 6.

If remainder of n/7 is 0, then

$remainder(\frac{n^2}{7})=(0\times 0)(\text{mod }7)=0$

If remainder of n/7 is 1, then

$remainder(\frac{n^2}{7})=(1\times 1)(\text{mod }7)=1$

If remainder of n/7 is 2, then

$remainder(\frac{n^2}{7})=(2\times 2)(\text{mod }7)=4$

If remainder of n/7 is 3, then

$remainder(\frac{n^2}{7})=(3\times 3)(\text{mod }7)=9(\text{mod }7)=2$

If remainder of n/7 is 4, then

$remainder(\frac{n^2}{7})=(4\times 4)(\text{mod }7)=16(\text{mod }7)=2$

If remainder of n/7 is 5, then

$remainder(\frac{n^2}{7})=(5\times 5)(\text{mod }7)=25(\text{mod }7)=4$

If remainder of n/7 is 6, then

$remainder(\frac{n^2}{7})=(6\times 6)(\text{mod }7)=36(\text{mod }7)=1$

For every n E N, when we divide n2 by 7, the remainder is either 0, 1, 2, or 4.

Hence proved.