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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.

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

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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.

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Just because $(\sqrt{n^2})^4=n^4$.

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  • $\begingroup$ and then why 0? $\endgroup$ – ahihelp Jun 24 '17 at 8:08
  • $\begingroup$ By definition, the remainder is zero if and only if the number divides the other one. In fact if you do $\frac{n^4}{n}=n^3+0$ $\endgroup$ – Alberto Andrenucci Jun 24 '17 at 8:21

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