# Could you please explain why the remainder when the square of 49 is divided by the square root of 49 is 0?

Here is the original question,

What is the remainder when the square of 49 is divided by the square root of 49?

According to UKMT,

Because 49 = 7×7, it follows both that $\sqrt 49$ = 7 and that 7 is a factor of 49.

There fore 7 is also a factor of $49^2$.

Hence the remainder when the square of 49 is divided by the square root of 49 is 0

They also gave the investigation,

1. Explain why it is true that for each positive integer n,

$(n^2)^2 ÷ \sqrt{n^2} =0$

2. Follow investigation 1, is the remainder still 0 in the case where n is a negative integer?

The answer I came up with was 343.

And I still don't get why it is true that the remainder is 0 in investigation 1.

Could you please explain it in more detail, and if it's okay, please also explain investigation 2.

Thank you for your help.

• Because $n$ is a factor of $n^4$ ? – samjoe Jun 24 '17 at 7:49
• @samjoe Could explain more please? – ahihelp Jun 24 '17 at 8:02
• Do you know the definition of a remainder? – Deepak Jun 24 '17 at 8:14
• $n^4=n^3\times n+0$ where the second term on RHS stands for the remainder when there is division by $n$. How did you get $343$? – drhab Jun 24 '17 at 8:15
• Don't say $(n^2)^2÷\sqrt{n^2}=0$. That's not what "the remainder is $0$" means. – Thorgott Jun 24 '17 at 8:40

I think you're confusing "remainder" with "quotient." If you divide $49^2$ by $7$ the quotient is $343$. The remainder is $0$.
I think you're confusing the terms quotient and remainder lets try with an example. Find the quotient and remainder when $23$ is divided by $7$ $$23=\color{red}3\cdot 7+\color{blue}2$$ Now the red part is the quotient while the blue is the remainder now $$7^4=\color{red}{343}\cdot 7+\color{blue}0$$ The red part is equal to $343$ which is the quotient while the blue part is equal to $0$ which is the remainder.
Just because $(\sqrt{n^2})^4=n^4$.
• By definition, the remainder is zero if and only if the number divides the other one. In fact if you do $\frac{n^4}{n}=n^3+0$ – Alberto Andrenucci Jun 24 '17 at 8:21