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# +ve Integer, divided by 4 and 9

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Senior Manager
Joined: 26 Mar 2008
Posts: 337
Location: Washington DC
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Kudos [?]: 53 [0], given: 4

+ve Integer, divided by 4 and 9 [#permalink]

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12 Aug 2009, 22:35
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78% (02:17) correct 22% (01:09) wrong based on 10 sessions

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Q> When a positive integer is divided by 4, the remainder is r; when divided by 9, the remainder is R. What is the greatest possible value of r^2+R?
A> 23,
B> 21,
C> 17,
D> 13,
E> 11

Manager
Joined: 25 Jul 2009
Posts: 116
Schools: NYU, NUS, ISB, DUKE, ROSS, DARDEN
Followers: 4

Kudos [?]: 205 [4] , given: 17

Re: +ve Integer, divided by 4 and 9 [#permalink]

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13 Aug 2009, 01:08
4
KUDOS
Q> When a positive integer is divided by 4, the remainder is r; when divided by 9, the remainder is R. What is the greatest possible value of r^2+R?
A> 23,
B> 21,
C> 17,
D> 13,
E> 11

When a number, say n, is divided by 4 the maximum remainder possible is 3 and when n is divided by 9 the maximum remainder possible is 8 => r=3 & R=8

An eg of n is 35

Thus Max(r^2+R) is attained when both r & R are maximum i.e. when r=3 & R=8
=>Max(r^2+R) = 3^2 + 8
=>Max(r^2+R) = 17

ANS: C
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Joined: 11 Aug 2009
Posts: 2
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Kudos [?]: 0 [0], given: 1

Re: +ve Integer, divided by 4 and 9 [#permalink]

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13 Aug 2009, 05:37
samrus98 wrote:
Q> When a positive integer is divided by 4, the remainder is r; when divided by 9, the remainder is R. What is the greatest possible value of r^2+R?
A> 23,
B> 21,
C> 17,
D> 13,
E> 11

When a number, say n, is divided by 4 the maximum remainder possible is 3 and when n is divided by 9 the maximum remainder possible is 8 => r=3 & R=8

An eg of n is 35

Thus Max(r^2+R) is attained when both r & R are maximum i.e. when r=3 & R=8
=>Max(r^2+R) = 3^2 + 8
=>Max(r^2+R) = 17

ANS: C

You gave one example of such number (35), is there a formula to generate all the numbers that satify this condition? I am asking because it might so happen in the exam that numbers and remainders are such that we are not able to think of sample number (35 in this case) that satifies the condition.......or is it that the numbers will always satisfy the condition?
SVP
Joined: 05 Jul 2006
Posts: 1512
Followers: 5

Kudos [?]: 212 [0], given: 39

Re: +ve Integer, divided by 4 and 9 [#permalink]

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13 Aug 2009, 09:27
samrus98 wrote:
Q> When a positive integer is divided by 4, the remainder is r; when divided by 9, the remainder is R. What is the greatest possible value of r^2+R?
A> 23,
B> 21,
C> 17,
D> 13,
E> 11

When a number, say n, is divided by 4 the maximum remainder possible is 3 and when n is divided by 9 the maximum remainder possible is 8 => r=3 & R=8

An eg of n is 35

Thus Max(r^2+R) is attained when both r & R are maximum i.e. when r=3 & R=8
=>Max(r^2+R) = 3^2 + 8
=>Max(r^2+R) = 17

ANS: C

excellent approach...text book type of answer..Kudos from me , thanks
Intern
Joined: 17 Nov 2009
Posts: 37
Schools: University of Toronto, Mcgill, Queens
Followers: 0

Kudos [?]: 82 [0], given: 9

Re: +ve Integer, divided by 4 and 9 [#permalink]

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13 Feb 2010, 00:47
samrus98 wrote:
Q> When a positive integer is divided by 4, the remainder is r; when divided by 9, the remainder is R. What is the greatest possible value of r^2+R?
A> 23,
B> 21,
C> 17,
D> 13,
E> 11

When a number, say n, is divided by 4 the maximum remainder possible is 3 and when n is divided by 9 the maximum remainder possible is 8 => r=3 & R=8

An eg of n is 35

Thus Max(r^2+R) is attained when both r & R are maximum i.e. when r=3 & R=8
=>Max(r^2+R) = 3^2 + 8
=>Max(r^2+R) = 17

ANS: C

Excellent approach.
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Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2797
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 219

Kudos [?]: 1452 [0], given: 235

Re: +ve Integer, divided by 4 and 9 [#permalink]

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13 Feb 2010, 01:46
sandil00 wrote:
samrus98 wrote:
Q> When a positive integer is divided by 4, the remainder is r; when divided by 9, the remainder is R. What is the greatest possible value of r^2+R?
A> 23,
B> 21,
C> 17,
D> 13,
E> 11

When a number, say n, is divided by 4 the maximum remainder possible is 3 and when n is divided by 9 the maximum remainder possible is 8 => r=3 & R=8

An eg of n is 35

Thus Max(r^2+R) is attained when both r & R are maximum i.e. when r=3 & R=8
=>Max(r^2+R) = 3^2 + 8
=>Max(r^2+R) = 17

ANS: C

You gave one example of such number (35), is there a formula to generate all the numbers that satify this condition? I am asking because it might so happen in the exam that numbers and remainders are such that we are not able to think of sample number (35 in this case) that satifies the condition.......or is it that the numbers will always satisfy the condition?

All those numbers will satisfy this when it is of type 4n-1 = 9m-1

=> 4n= 9m this happens when n=9 and m =4 thus 4n-1=35
take n =18 and m =8 thus 4n-1 = 71 and so on.

I hope this helps.
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Senior Manager
Joined: 01 Feb 2010
Posts: 267
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Kudos [?]: 47 [0], given: 2

Re: +ve Integer, divided by 4 and 9 [#permalink]

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13 Feb 2010, 11:07
marshpa wrote:
Q> When a positive integer is divided by 4, the remainder is r; when divided by 9, the remainder is R. What is the greatest possible value of r^2+R?
A> 23,
B> 21,
C> 17,
D> 13,
E> 11

r can be 1,2 or 3.
R can be 1,2,3 ... 8

maximum of r^2 + R can be 3^2 + 8 = 17 hence C.
Manager
Joined: 20 Apr 2010
Posts: 210
Schools: ISB, HEC, Said
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Kudos [?]: 61 [0], given: 28

Re: +ve Integer, divided by 4 and 9 [#permalink]

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15 Sep 2010, 07:01
Excellent explaination given by samrus98 Kudos to you
Re: +ve Integer, divided by 4 and 9   [#permalink] 15 Sep 2010, 07:01
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