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

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New post Updated on: 29 Oct 2018, 04:00
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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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Originally posted by EgmatQuantExpert on 24 Oct 2018, 06:01.
Last edited by EgmatQuantExpert on 29 Oct 2018, 04:00, edited 8 times in total.
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Re: The number A can be expressed as p x q, where p and q are positive int  [#permalink]

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New post 24 Oct 2018, 07:06
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
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Re: The number A can be expressed as p x q, where p and q are positive int  [#permalink]

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New post 30 Oct 2018, 05:25
1

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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Re: The number A can be expressed as p x q, where p and q are positive int  [#permalink]

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New post 30 Oct 2018, 13:13
EgmatQuantExpert wrote:

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. \(q^2 – 8q + 15 = 0\)

\(p,q\,\, \geqslant 1\,\,\,{\text{ints}}\)

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

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

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

\(\left( {1 + 2} \right)\,\,\,pq = {2^3} \cdot M\,,\,\,M\,\,{\text{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle\)


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

Regards,
Fabio.
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Re: The number A can be expressed as p x q, where p and q are positive int &nbs [#permalink] 30 Oct 2018, 13:13
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