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When a positive integer is divided by 4, the remainder is r; when divi [#permalink]
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28 Dec 2016, 01:22
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Re: When a positive integer is divided by 4, the remainder is r; when divi [#permalink]
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28 Dec 2016, 02:16
Option C)Given: r = Remainder[\(\frac{N}{4}\)] & R = Remainder[\(\frac{N}{9}\)] : Greatest possible value of (r^2 + R) ? N = K*LCM(4,9) + X i.e., N = 36 + X or 72 + X or . . . To find Greatest possible value of (r^2 + R): we need to maximize r and R. For a given number M, greatest possible remainder could be (M1). Similarly, greatest possible remainder for N could be ( LCM(4,9)  1) = 35 So, N could take values = 36 + 35 or 72 + 35 or . . . And, r = 3 & R = 8. Here, r = 3 (i.e., 41 = greatest possible remainder a number can have on dividing by 4) and R = 8 (i.e., 91 = greatest possible remainder a number can have on dividing by 9) Hence, Greatest possible value of ( r^2 + R) = ( 3^2 + 8) = 17
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When a positive integer is divided by 4, the remainder is r; when divi [#permalink]
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30 Dec 2016, 23:30
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Bunuel wrote: 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 SolutionWhen a positive integer is divided by 4, the remainder can be 0,1,2 or 3. Hence we can infer that the value of r can be 0,1,2 or 3 . Similarly, when a positive integer is divided by 9, the remainder can be 0,1,2,3....8 Hence we can infer that the value of R can be 0,1,2,3,4,5,6,7 or 8. Since we have to maximize \(r^2 + R\), we need to consider the maximum value of both r and R, which will be 3 and 8 respectively. Hence the maximum value of the expression is: \(r^2 + R\) \(=3^2 + 8\) \(=9 + 8\) \(=17\) Correct Answer : Option CThanks, Saquib Quant Expert eGMATTo practise ten 700+ Level Number Properties Questions attempt the The EGMAT Number Properties Knockout
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Re: When a positive integer is divided by 4, the remainder is r; when divi [#permalink]
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04 Jan 2017, 21:37
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\(n=4p+r\) \(n=9q+R\)
the remainder is always smaller then the dividor, so r<4, and R<9. Since the remainder is always an integer, the greatest possible value for r and R is: r=3, R=8
\(r^2+R=9+8=17\)



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Re: When a positive integer is divided by 4, the remainder is r; when divi [#permalink]
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04 Jan 2017, 22:21
Bunuel wrote: 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 Using the concept of divisibility, you can get the answer with minimum calculation. When a number is divided by 4, the maximum value of r can be 3. When a number is divided by 9, the maximum value of R can be 8. Now all we need to figure out is whether we can have a number such that when divided by 4, it gives remainder 3 and when divided by 9, it gives remainder 8. Imagine the number split into groups of 9 and 8 leftover. When the same is divided by 4, the 8 leftover will be evenly split into groups of 4. From each group of 9, we will make 2 groups of 4 each such that 1 is leftover from each group of 9. We want 3 to be leftover when we divide by 4 and this will be possible if we have 3 groups of 9. So basically such a number could be 3*9 + 8 = 35 etc Hence, maximum value of r^2 + R = 3^2 + 8 = 17 Answer (C)
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Re: When a positive integer is divided by 4, the remainder is r; when divi [#permalink]
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29 Jan 2018, 03:44
Is this a real 600 level question? :O



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Re: When a positive integer is divided by 4, the remainder is r; when divi [#permalink]
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29 Jan 2018, 05:27
MvArrow wrote: Is this a real 600 level question? :O No. About 680  700 level I would say.
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Re: When a positive integer is divided by 4, the remainder is r; when divi [#permalink]
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29 Jan 2018, 05:28



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When a positive integer is divided by 4, the remainder is r; when divi [#permalink]
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29 Jan 2018, 08:00
Bunuel wrote: 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 Another way that works is to gauge the LCM of the two divisors (4 and 9) and the LCM  1 will yield that number (\(361=35\)) for which both divisions result in the maximum remainder (3 and 8 respectively). I don´t actually know why this works (probably is based on the theory that some of the other users have already posted) but if you encounter something similar, it´s a readytoapply and go. 
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When a positive integer is divided by 4, the remainder is r; when divi
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