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When a positive integer is divided by 4, the remainder is r; when divi

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When a positive integer is divided by 4, the remainder is r; when divi [#permalink]

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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
[Reveal] Spoiler: OA

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Re: When a positive integer is divided by 4, the remainder is r; when divi [#permalink]

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New post 28 Dec 2016, 01:16
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 (M-1).
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., 4-1 = greatest possible remainder a number can have on dividing by 4) and R = 8 (i.e., 9-1 = 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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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


Solution


When 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 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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\(n=4p+r\)
\(n=9q+R\)

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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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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New post 29 Jan 2018, 02:44
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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New post 29 Jan 2018, 04:27
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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New post 29 Jan 2018, 04:28
MvArrow wrote:
Is this a real 600 level question? :O


You can check difficulty level of a question along with the stats on it in the first post. For this question Difficulty: 600-700 Level. The difficulty level of a question is calculated automatically based on the timer stats from the users which attempted the question.
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Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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When a positive integer is divided by 4, the remainder is r; when divi [#permalink]

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New post 29 Jan 2018, 07:00
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 (\(36-1=35\)) 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 ready-to-apply and go.


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When a positive integer is divided by 4, the remainder is r; when divi   [#permalink] 29 Jan 2018, 07:00
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