[Math] Could you please explain why the remainder when the square of 49 is divided by the square root of 49 is 0

algebra-precalculus

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.

Best Answer

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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