Problem of the Day


Let m, a, and r represent positive integers such that \sqrt{a}<r<\sqrt{2a}. If r is the remainder when m is divided by a, what is the remainder when m^2 is divided by a?

A. r
B. r^2
C. r^2-a
D. \sqrt{r}
E. 2r

Reveal Answer

Answer

[latexpage]

C. $r^2-a$

See the Solution

Solution

[latexpage]

We know that $m$ divided by $a$ leaves a remainder of $r$. This means that $m=an+r$ for some integer, $n$. Squaring $m$ gives us:

$m^2=(an+r)^2=(an+r)(an+r)=a^2n^2+2anr+r^2$.

We can write this expression as $m^2=a(an^2+2nr)+r^2$. This seems to imply that $m^2$ divided by $a$ would have a quotient of $an^2+2nr$ and a remainder of $r^2$. The only problem is that $r^2$ may be greater than or equal to $a$, and this wouldn’t fit the mathematical definition of a remainder. To see this clearly, consider some real numbers. $30$ divided by $7$ is $4$, remainder $2$. If we said $30$ divided by $7$ is $3$, remainder $9$, the math is okay in a sense, but we could have divided $7$ into $30$ one more time, hence the typical restriction that a remainder be less than the divisor.

In this case, we know that $\sqrt{a}<r<\sqrt{2a}$. All of these numbers are positive, so we can square all of them and still preserve the inequality: $a<r^2<2a$. $r^2$ is bigger than $a$, but less than $2a$ which means we could have divided $a$ into $r^2$ exactly one more time, leaving a remainder of $r^2-a$.

 

 


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