The number A can be expressed as p x q, where p and q are positive int

Question

Common Mistakes One Must Avoid in Remainders – Practice question 1

The number A can be expressed as p x q, where p and q are positive integers. Is A divisible by 16?
 1. p = 8 x k, where k is an odd number.
2.  q^2 – 8q + 15 = 0

A)  Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
 B)  Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
 C)  BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
 D)  EACH statement ALONE is sufficient.
 E)  Statements (1) and (2) TOGETHER are NOT sufficient.

To solve question 2:    Question 2

To read the article:      Common Mistakes One Must Avoid in Remainders

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doomedcat · Answer

The number A can be expressed as p x q, where p and q are positive integers. Is A divisible by 16?
1. p = 8 x k, where k is an odd number.
2. q2–8q+15=0

Stat1) p = 8 x k, where k is an odd number: Now the A can be written as  \frac{8k*q}{16} 
 Now if q is multiple of 2 then A is divisible by 16 else not, Remember k cannot be multiple of 2 as k is odd, Hence Not suff

Stat2)  q^2–8q+15=0  , solving we get q to be either 5 or 3; now A becomes  \frac{p*5}{16}  or  \frac{p*3}{16} 
 if p is multiple of 16 then A is divisible by 16 else not , hence Not suff

Both 1 & 2 )  \frac{8k*5}{16}  or  \frac{8k*3}{16}  ==> as k, q is odd and 8 is not divisible by 16 so A cannot be divisible by 16 , Hence Suff C

EgmatQuantExpert · Answer

Solution 
  Given:  
 • The number A can be expressed as p x q.
• Both p and q are positive integers.

To find:  
 • Whether A is divisible by 16 or not.

Analysing Statement 1  
As per the information given in statement 1, p = 8 x k, where k is an odd number.

Now if A is a multiple of 16, it must be a multiple of both 8 and 2 separately.
 • From statement 1, we only know A is a multiple of 8 but not 16.
• Also, we have no information about q – whether q is an even or odd number.

Therefore, we cannot determine whether A is divisible by 16 or not.

Hence, statement 1 is not sufficient to answer the question.

Analysing Statement 2  
As per the information given in statement 2,  q^2 – 8q + 15 = 0

Simplifying the equation, we can write,
 •  q^2 – 8q + 15 = 0 
Or,  q^2 – 5q – 3q + 15 = 0 
Or, q (q – 5) – 3 (q – 5) = 0
Or, (q – 5) (q – 3) = 0
Or, q = 3 or 5

From this statement, we can find possible values of q but have no information about p.

Hence, statement 2 is not sufficient to answer the question.

Combining Both Statements  
 • From statement 1, we know p = 8 x odd number.
• From statement 2, we know q = 3 or 5 = odd number
• Hence, A = p x q = 8 x odd number x odd number, which can never be a multiple of 16.

Therefore, combining the statements, we can say A is not divisible by 16.

Hence, the correct answer is option C.

Answer: C

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fskilnik · Answer

p,q\,\, \geqslant 1\,\,\,{	ext{ints}}

\frac{{pq}}{{{2^4}}}\,\,\mathop = \limits^? \,\,\operatorname{int}

\left( 1 ight)\,\,\,p = 8k\,\,,\,\,\,k\,\,{	ext{odd}}\,\,\,\,\left\{ \begin{gathered}
 \,{	ext{Take}}\,\,\left( {p,q} ight) = \left( {8,1} ight)\,\,\,\, \Rightarrow \,\,\,\left\langle {{	ext{NO}}} ightangle \,\, \hfill \
 \,{	ext{Take}}\,\,\left( {p,q} ight) = \left( {8,2} ight)\,\,\,\, \Rightarrow \,\,\,\left\langle {{	ext{YES}}} ightangle \,\, \hfill \ 
\end{gathered} ight.

\left( 2 ight)\,\,\,\left( {q - 3} ight)\left( {q - 5} ight) = 0\,\,\,\,\, \Rightarrow \,\,\,\,\,q = 3\,\,\,{	ext{or}}\,\,\,q = 5\,\,\,\,\,\left\{ \begin{gathered}
 \,{	ext{Take}}\,\,\left( {p,q} ight) = \left( {8,3} ight)\,\,\,\, \Rightarrow \,\,\,\left\langle {{	ext{NO}}} ightangle \,\, \hfill \
 \,{	ext{Take}}\,\,\left( {p,q} ight) = \left( {16,3} ight)\,\,\,\, \Rightarrow \,\,\,\left\langle {{	ext{YES}}} ightangle \,\, \hfill \ 
\end{gathered} ight.

\left( {1 + 2} ight)\,\,\,pq = {2^3} \cdot M\,,\,\,M\,\,{	ext{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{	ext{NO}}} ightangle

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

Regards, 
Fabio.

MathRevolution · Answer

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. Visit https://www.mathrevolution.com/gmat/lesson for details. Since we have 3 variables ( A, p and q ) and 1 equation( A=p \cdot q ), C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first. Conditions 1) & 2) Condition 2 tells q = 3 or q = 5 since q^2-8q+15 = (q-3)(q-5)= 0 . Then we have A = 8 \cdot 3 \cdot k or A = 5 \cdot 8 \cdot k for some odd integer k . A is not divisible by 16 and the answer is always 'no'. Since both conditions together yield a unique solution, they are sufficient. Since 'no' is also a unique answer by CMT (Common Mistake Type) 1, both conditions are sufficient, when used together. Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B. Condition 1) If p = 8 and q = 2 , then A = p \cdot q = 16 is divisible by 16 and the answer is 'yes'. If p = 8 and q = 3 , then A = p \cdot q = 24 is not divisible by 16 and the answer is 'no'. Since condition 1) does not yield a unique solution, it is not sufficient Condition 2) If p = 16 and q = 3 , then A = p \cdot q = 32 is divisible by 16 and the answer is 'yes'. If p = 8 and q = 5 , then A = p \cdot q = 40 is not divisible by 16 and the answer is 'no'. Since condition 2) does not yield a unique solution, it is not sufficient Therefore, C is the answer. Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.

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The number A can be expressed as p x q, where p and q are positive int

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