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# +ve Integer, divided by 4 and 9

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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

I have doubt on question. Please post answer with explanation.
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Re: +ve Integer, divided by 4 and 9 [#permalink]

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13 Aug 2009, 00:08
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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

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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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?
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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
Intern
Joined: 17 Nov 2009
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Kudos [?]: 96 [0], given: 9

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 [?]: 1617 [0], given: 235

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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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

I have doubt on question. Please post answer with explanation.

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.
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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
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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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# +ve Integer, divided by 4 and 9

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