# Positive integer not divisible by 7

• Jan 14th 2011, 07:02 AM
Hellbent
Positive integer not divisible by 7
Hi,

Is there a simple explanation for this?

The positive integer $n$ is not divisible by $7$. The remainder when $n^2$ is divided by 7 and the remainder when $n$ is divided by $7$ are each equal to $k$. What is $k$?

A. 1
B. 2
C. 4
D. 6
E. It cannot be determined from the information given.

I just looked at the middle part and concluded the answer to be 1. The only numbers I know that are equal to their squares are 0 and 1. Zero isn't available.
• Jan 14th 2011, 07:18 AM
Unknown008
I can't see any answer besides A. ..

Note that it cannot be zero because otherwise, n would be a multiple of 7.
• Jan 14th 2011, 07:27 AM
Pranas
Quote:

Originally Posted by Hellbent
Hi,

Is there a simple explanation for this?

The positive integer $n$ is not divisible by $7$. The remainder when $n^2$ is divided by 7 and the remainder when $n$ is divided by $7$ are each equal to $k$. What is $k$?

A. 1
B. 2
C. 4
D. 6
E. It cannot be determined from the information given.

I just looked at the middle part and concluded the answer to be 1. The only numbers I know that are equal to their squares are 0 and 1. Zero isn't available.

I have to agree.
Let's say our number can be written as:
$$n = 7m + k$$
Therefore $${n^2} = 49 \cdot {m^2} + 14 \cdot m \cdot k + {k^2} = 7 \cdot (7 \cdot {m^2} + 2 \cdot m \cdot k) + {k^2}$$

We actually have to keep in mind, that we are trying to evaluate $$k$$ itself and only remainder of $${k^2}$$ when it's divided by 7.
However, in your case there really are no other possible scenarios other than $$k = {k^2} = 1$$ (you can run through $$k = \overline {1,6}$$ to make sure of that).
• Jan 15th 2011, 04:46 AM
Quote:

Originally Posted by Hellbent
Hi,

Is there a simple explanation for this?

The positive integer $n$ is not divisible by $7$. The remainder when $n^2$ is divided by 7 and the remainder when $n$ is divided by $7$ are each equal to $k$. What is $k$?

A. 1
B. 2
C. 4
D. 6
E. It cannot be determined from the information given.

I just looked at the middle part and concluded the answer to be 1. The only numbers I know that are equal to their squares are 0 and 1. Zero isn't available.

$\displaystyle\frac{n^2}{7}=m+\frac{k}{7}\Rightarro w\ n^2-7m-k=0$

$\displaystyle\frac{n}{7}=x+\frac{k}{7}\Rightarrow\ n-7x=k$

Substitute k

$n^2-7m-n+7x=0\Rightarrow\ n^2-n+7(x-m)=0$

$\Rightarrow\ (n+7)(n+x-m)=n^2+(-1)n+7(x-m)\Rightarrow\ x-m=-8\Rightarrow\ m-x=8$

$(n-8)(n+7)=0\Rightarrow\ n=8$

$\displaystyle\frac{n^2}{7}=\frac{64}{7}=9\frac{1}{ 7}$

$\displaystyle\frac{8}{7}=1\frac{1}{7}$

The remainder is 1.
• Jan 15th 2011, 05:37 AM
Pranas
Quote:

$\displaystyle\frac{n^2}{7}=m+\frac{k}{7}\Rightarro w\ n^2-7m-k=0$

$\displaystyle\frac{n}{7}=x+\frac{k}{7}\Rightarrow\ n-7x=k$

Substitute k

$n^2-7m-n+7x=0\Rightarrow\ n^2-n+7(x-m)=0$

$\Rightarrow\ (n+7)(n+x-m)=n^2+(-1)n+7(x-m)\Rightarrow\ x-m=-8\Rightarrow\ m-x=8$

$(n-8)(n+7)=0\Rightarrow\ n=8$

$\displaystyle\frac{n^2}{7}=\frac{64}{7}=9\frac{1}{ 7}$

$\displaystyle\frac{8}{7}=1\frac{1}{7}$

The remainder is 1.

Could you please clarify how did you come up with
$\displaystyle \((n+7)(n+x-m)=n^2+(-1)n+7(x-m)\Rightarrow\ x-m=-8\Rightarrow\ m-x=8$

and $\displaystyle ${\rm{n = 8}}$$

If you concluded, that there is only one appropriate value for $\displaystyle ${n}}$$ under given circumstances, that is not true.
Otherwise my apologies if I misunderstood something.
• Jan 15th 2011, 05:51 AM
$\displaystyle \((n+7)(n+x-m)=n^2+(-1)n+7(x-m)\Rightarrow\ x-m=-8\Rightarrow\ m-x=8$
and $\displaystyle ${\rm{n = 8}}$$
If you concluded, that there is only one appropriate value for $\displaystyle ${n}}$$ under given circumstances, that is not true.