$\begingroup$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.
$\endgroup$53 Answers
$\begingroup$I think you're confusing "remainder" with "quotient." If you divide $49^2$ by $7$ the quotient is $343$. The remainder is $0$.
$\endgroup$$\begingroup$I think you're confusing the terms quotient and remainder lets try with an example. Find the quotient and remainder when $23$ is divided by $7$ $$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$$ The red part is equal to $343$ which is the quotient while the blue part is equal to $0$ which is the remainder.
$\endgroup$$\begingroup$Just because $(\sqrt{n^2})^4=n^4$.
$\endgroup$2
