What is the remainder when a positive integer ‘P’ is divided by 33?

Question

(1) When P is divided by 11, the remainder is 5.
(2) When P is divided by 99, the remainder is 5.

BrushMyQuant · Accepted Answer

I have written a detailed post on  How to Solve Remainder Problems . Please go through it to understand the topic in bit more detail.

STAT1 : When P is divided by 11, the remainder is 5 
Now P can be taken as P = 11k + 5 ( Dividend = Divisor * Quotient + Remainder )

Now when 11k+5 (which is P) is divided by 33 then we can get multiple remainders depending on the value of k
Example
k = 0 then P = 5 and P/33 remainder is 5
k = 1 then P = 16 and P/33 remainder is 16
k = 2 then P = 27 and P/33 remainder is 27
Since we get multiple values for remainders so NOT SUFFICIENT

STAT2: When P is divided by 99, the remainder is 5 
As in STAT1, P = 99t + 5
Now, when 99t + 5 (which is P) is dividend by 33 then we get a remainder of 5 ( 0 remainder from 99t and 5 remainder from 5 => total remainder of 5)
Since, we get only one value of remainder in this case so it is SUFFICIENT

So, Answer will be B
Hope it helps!

Watch the following video to learn the Basics of Remainders

https://www.youtube.com/watch?v=P0S6j2uTaj4

fskilnik · Answer

P \ge 1\,\,{\mathop{\rm int}} \,\,\left( * \right)

P = 33M + R

M\mathop \ge \limits^{\left( * \right)} 0\,\,{\mathop{\rm int}} \,\,\,,\,\,\,0 \le R\,\,{\mathop{\rm int}} \,\, \le 32

? = R

\left( 1 \right)\,\,P = 11K + 5\,\,\,,\,\,\,K\mathop \ge \limits^{\left( * \right)} 0\,\,{\mathop{\rm int}} \,\,\,\,\left\{ \matrix{
 \,{\rm{Take}}\,\,K = 0\,\,\,\, \Rightarrow \,\,\,R = 5\,\, \hfill \cr 
 \,{\rm{Take}}\,\,K = 1\,\,\,\, \Rightarrow \,\,\,R = 16\, \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{INSUFF}}.

\left( 2 \right)\,\,P = 99L + 5\,\,\,,\,\,\,L\mathop \ge \limits^{\left( * \right)} 0\,\,{\mathop{\rm int}} \,\,\,\,\, \Rightarrow \,\,\,\,\,P = 33\left( {3L} \right) + 5\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{
 \,M = 3L \hfill \cr 
 \,? = R = 5\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}{\rm{.}} \hfill \cr} \right.

This solution follows the notations and rationale taught in the GMATH method.

Regards, 
Fabio.

KanishkM · Answer

Remainder question, cool

From 1 
P/11, R =5

P = 11N +5
N = 0, P =5
N = 1, P = 17 
N = 2, P = 27
When put back in the question will give us remainders as 5,17 and 27
Insufficient

From 2 
P/99, R = 5

P = 99N + 5
N = 0, P =5
N = 1, P = 104
N = 2, P = 203
When put back in the question will give us remainder as only 5
Sufficient

B

lucasd14 · Answer

Statement 1:  
P = 11u + 5
P can take several values prior and after reaching 33, with different remainders.
P= 5, 16, ... Just checking the first two possibilities, we see they give different remainders (5 and 16, each).
 Not Sufficient.

Statement 2:  
P = 99v + 5
We can see that 99 is multiple of 33. Thus, remainder will be 5 in every case.
 Sufficient.

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What is the remainder when a positive integer ‘P’ is divided by 33?

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