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705-805 Level|   Remainders|      
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Let m, a, and r represent positive integers such that a^(1/2) < r < 2a 27 Nov 2019, 02:12
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58% (02:33) correct 42% (02:19) wrong based on 219 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. \(2r\)


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C
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GMAT Focus 1: 735 Q90 V89 DI81
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Let m, a, and r represent positive integers such that a^(1/2) < r < 2a 27 Nov 2019, 04:59
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Bunuel
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. \(2r\)


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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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Let m, a, and r represent positive integers such that a^(1/2) < r < 2a 27 Nov 2019, 09:12
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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

so from given answer options
\(r^2-a\)
\(3^2-8=1\)

IMO C

Bunuel
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. \(2r\)


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Let m, a, and r represent positive integers such that a^(1/2) < r < 2a 28 Nov 2019, 04:23
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Bunuel
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. \(2r\)


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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:)
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Let m, a, and r represent positive integers such that a^(1/2) < r < 2a 07 Mar 2025, 11:02
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This is a good question. So it was easy to deduce till that since the remainder is r. When m^2 is divided by a then the remainder will r^2.

But here comes the questions if r^2 is greater than a then the final remainder would be less than r^2.

This is where the given expression helps. Now it is given that, √a<r<√2a, if you square it, we get a<r^2<2a. It took me so much time to understand what this means.

Since, r^2 is greater than a, you would need to minus a to get the final remainder. But since it is less than 2a, we would not need to do it again.

For example -
11/4 gives remainder = 3
but 121 (11^2) when divided by 4 gives you the remainder of 1. Which you see is like 3*3 = 9 - 4 - 4 = 1.

Hence, since r^2 is less than 2a, we don't need to minus that again which gives the final answer as r^2 - a.

Hope this helps.
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