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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Let m, a, and r represent positive integers such that a^(1/2) < r < 2a

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Math Expert V
Joined: 02 Sep 2009
Posts: 59721
Let m, a, and r represent positive integers such that a^(1/2) < r < 2a  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 54% (02:25) correct 46% (02:07) wrong based on 24 sessions

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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. $$2r$$

Are You Up For the Challenge: 700 Level Questions

_________________
Math Expert V
Joined: 02 Aug 2009
Posts: 8310
Re: Let m, a, and r represent positive integers such that a^(1/2) < r < 2a  [#permalink]

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Bunuel wrote:
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. $$2r$$

Are You Up For the Challenge: 700 Level Questions

SUBSTITUTION
Easiest way would be to take values..

$$\sqrt{a}<r<\sqrt{2a}$$.....$$\sqrt{9}<r<\sqrt{2*9}.....3<4<3\sqrt{2}$$

Now m=9*x+4= 9*1+4=13..
13 divided by 9 gives 4 as reaminder..
13^2 divided by 9....169=9*18+7, so remainder =7..

Check the choice that gives 7..

A. $$r......4$$...NO

B. $$r^2....16$$...NO

C. $$r^2-a....16-9=7$$...Yes

D. $$\sqrt{r}....2$$...NO

E. $$2r......284=8$$...NO

ALGEBRAIC way..
m divided by a gives r as the remainder, so m^2 will give r^2...
BUT r^2 will be greater than a, since $$\sqrt{a}<r<\sqrt{2a}....a<r^2<2a$$, so remainder will be $$r^2-a$$
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GMAT Club Legend  V
Joined: 18 Aug 2017
Posts: 5483
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: Let m, a, and r represent positive integers such that a^(1/2) < r < 2a  [#permalink]

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let a= 8 , r= 3
so
$$\sqrt{8}<3<\sqrt{28}$$
we get
2<3<4

given condition ; if r is the remainder when m is divided by a, so m = 5 and a = 8 ; r=3
now
what is the remainder when m^2 is divided by a?
25/8 ; remainder = 1

$$r^2-a$$
$$3^2-8=1$$

IMO C

Bunuel wrote:
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. $$2r$$

Are You Up For the Challenge: 700 Level Questions
Manager  S
Joined: 10 Dec 2017
Posts: 151
Location: India
Let m, a, and r represent positive integers such that a^(1/2) < r < 2a  [#permalink]

### Show Tags

Bunuel wrote:
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. $$2r$$

Are You Up For the Challenge: 700 Level Questions

Since r is a positive integer
$$a<r^2<2a$$
as a and r are positive integers and a>r(remainder is always less than the divisor)
a=3
$$3<r^2<6$$
$$r^2=4$$
r=2
a=3
so m=2
$$m^2=4$$
$$m^2/a$$gives remainder 1
$$r^2-a$$=4-3=1
C:) Let m, a, and r represent positive integers such that a^(1/2) < r < 2a   [#permalink] 28 Nov 2019, 04:23
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