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

Question

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

EgmatQuantExpert · Accepted Answer

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

Thanks,
Saquib
Quant Expert
e-GMAT

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RR88 · Answer

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

louwi · Answer

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

KarishmaB · Answer

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)

MvArrow · Answer

Is this a real 600 level question? :O

KarishmaB · Answer

No. About 680 - 700 level I would say.

Bunuel · Answer

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.

RooIgle · Answer

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

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