GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 13 Dec 2019, 13:55

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Let m, a, and r represent positive integers such that a^(1/2) < r < 2a

Author Message
TAGS:

### Hide Tags

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

### Show Tags

27 Nov 2019, 02:12
00:00

Difficulty:

55% (hard)

Question Stats:

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

### HideShow timer Statistics

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

### Show Tags

27 Nov 2019, 04:59
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 $$r^2-a$$
_________________
GMAT Club Legend
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]

### Show Tags

27 Nov 2019, 09:12
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
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

28 Nov 2019, 04:23
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
Display posts from previous: Sort by