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# +ve Integer, divided by 4 and 9

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Senior Manager
Joined: 26 Mar 2008
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+ve Integer, divided by 4 and 9 [#permalink]

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12 Aug 2009, 21:35
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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

Kudos [?]: 94 [0], given: 4

Manager
Joined: 25 Jul 2009
Posts: 114

Kudos [?]: 266 [4], given: 17

Schools: NYU, NUS, ISB, DUKE, ROSS, DARDEN
Re: +ve Integer, divided by 4 and 9 [#permalink]

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13 Aug 2009, 00:08
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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Kudos [?]: 266 [4], given: 17

Intern
Joined: 11 Aug 2009
Posts: 2

Kudos [?]: [0], given: 1

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

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13 Aug 2009, 04:37
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?

Kudos [?]: [0], given: 1

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Joined: 05 Jul 2006
Posts: 1748

Kudos [?]: 447 [0], given: 49

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

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13 Aug 2009, 08:27
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

Kudos [?]: 447 [0], given: 49

Intern
Joined: 17 Nov 2009
Posts: 36

Kudos [?]: 118 [0], given: 9

Schools: University of Toronto, Mcgill, Queens
Re: +ve Integer, divided by 4 and 9 [#permalink]

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12 Feb 2010, 23:47
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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Kudos [?]: 118 [0], given: 9

CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2754

Kudos [?]: 1926 [0], given: 235

Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Re: +ve Integer, divided by 4 and 9 [#permalink]

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13 Feb 2010, 00:46
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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Kudos [?]: 1926 [0], given: 235

Senior Manager
Joined: 01 Feb 2010
Posts: 251

Kudos [?]: 62 [0], given: 2

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

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13 Feb 2010, 10:07
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.

Kudos [?]: 62 [0], given: 2

Manager
Joined: 19 Apr 2010
Posts: 207

Kudos [?]: 90 [0], given: 28

Schools: ISB, HEC, Said
Re: +ve Integer, divided by 4 and 9 [#permalink]

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15 Sep 2010, 06:01
Excellent explaination given by samrus98 Kudos to you

Kudos [?]: 90 [0], given: 28

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

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27 Dec 2016, 17:33
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Re: +ve Integer, divided by 4 and 9   [#permalink] 27 Dec 2016, 17:33
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