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P is a two-digit integer, which can be written as 30a + b, where a and

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P is a two-digit integer, which can be written as 30a + b, where a and  [#permalink]

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New post Updated on: 21 Nov 2018, 03:57
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Question Stats:

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Common Mistakes One Must Avoid in Remainders – Practice question 2

P is a two-digit integer, which can be written as 30a + b, where a and b are positive integers. Find the remainder, when P is divided by 3.

    (1) a = 3
    (2) \(b^3 – 5b^2 – 14b = 0\)


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 3: Question 3

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

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Originally posted by EgmatQuantExpert on 24 Oct 2018, 07:03.
Last edited by EgmatQuantExpert on 21 Nov 2018, 03:57, edited 7 times in total.
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Re: P is a two-digit integer, which can be written as 30a + b, where a and  [#permalink]

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New post 24 Oct 2018, 08:32
P is a two-digit integer, which can be written as 30a + b, where a and b are positive integers. Find the remainder, when P is divided by 3.

30a+b.... Now 30a will always be divisible by 3, so it will depend on b.

(1) a =4
Not required. Nothing about b
Insufficient

(2) \(b^3 – 5b^2 – 14b = 0\)
\(b^3 – 5b^2 – 14b = 0..........b(b^2-5b-14)=0.........b(b^2-7b+2b-14)=0........b(b-7)(b+2)=0\), so b can be -2, 0 or 7...
But b is a positive integer so ONLY possible value 7..
30a+7 divided by 3 will leave 1 as remainder
Sufficient

B
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Re: P is a two-digit integer, which can be written as 30a + b, where a and  [#permalink]

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New post Updated on: 24 Oct 2018, 08:34
P is a two-digit integer, which can be written as 30a + b, where a and b are positive integers. Find the remainder, when P is divided by 3.

(1) a =4
(2) b3–5b2–14b=0

Given : P=30a + b ; divide by 3
\(\frac{P}{3}\) = \(\frac{30a}{3}\) + \(\frac{b}{3}\)
Now, 30 is divisible by 3, hence 30a will leave a remainder 0 upon division by 3; so actual remainder will be given by value of \(\frac{b}{3}\)
Question now becomes b=?

Stat1) a=4 , gives nothing about b, hence Not suff

Stat2) \(b^3–5b^2–14b=0\), take common b out => \(b*(b^2-5b-14)=0\) so either b=0 or \(b^2-5b-14=0\) ; further solving the later we get b= 7,-2. Since it is given that b is positive so in all we get b is 7; So remainder will be 1 Hence Suff
B

Edited : Corrected my mistake of missing that b was positive so 0 wouldnt count anyway .. silly me
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Originally posted by doomedcat on 24 Oct 2018, 08:19.
Last edited by doomedcat on 24 Oct 2018, 08:34, edited 1 time in total.
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Re: P is a two-digit integer, which can be written as 30a + b, where a and  [#permalink]

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New post 24 Oct 2018, 08:34
doomedcat wrote:
P is a two-digit integer, which can be written as 30a + b, where a and b are positive integers. Find the remainder, when P is divided by 3.

(1) a =4
(2) b3–5b2–14b=0

Given : P=30a + b ; divide by 3
\(\frac{P}{3}\) = \(\frac{30a}{3}\) + \(\frac{b}{3}\)
Now, 30 is divisible by 3, hence 30a will leave a remainder 0 upon division by 3; so actual remainder will be given by value of \(\frac{b}{3}\)
Question now becomes b=?

Stat1) a=4 , gives nothing about b, hence Not suff

Stat2) \(b^3–5b^2–14b=0\), take common b out => \(b*(b^2-5b-14)=0\) so either b=0 or \(b^2-5b-14=0\) ; further solving the later we get b= 7,-2. Since it is given that b is positive so in all we get b is either 0 or 7; Now multiple values of b hence Not suff

Both 1&2) still no help

E



You have solved it well but have missed out on b as ONLY a positive integer and 0 is neither positive nor negative, so only 7 left..
Be careful on these wordings
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Re: P is a two-digit integer, which can be written as 30a + b, where a and  [#permalink]

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New post 24 Oct 2018, 22:01
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EgmatQuantExpert wrote:
Common Mistakes One Must Avoid in Remainders – Practice question 2

P is a two-digit integer, which can be written as 30a + b, where a and b are positive integers. Find the remainder, when P is divided by 3.

    (1) a =4
    (2) \(b^3 – 5b^2 – 14b = 0\)


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 read the article: Common Mistakes One Must Avoid in Remainders

Image


I understand answer has to be B.
I just want to understand question; As it is given P is a two digit integer written as 30a+b, where a and b is positive integer. Then option how can be the option A) a=4 is valid?

As per Gmat DS options are always true.

Please clarify. Thanks


Sent from my iPad using GMAT Club Forum mobile app
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Re: P is a two-digit integer, which can be written as 30a + b, where a and  [#permalink]

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New post 30 Oct 2018, 06:13
vishalkazone wrote:
EgmatQuantExpert wrote:
Common Mistakes One Must Avoid in Remainders – Practice question 2

P is a two-digit integer, which can be written as 30a + b, where a and b are positive integers. Find the remainder, when P is divided by 3.

    (1) a =4
    (2) \(b^3 – 5b^2 – 14b = 0\)


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 read the article: Common Mistakes One Must Avoid in Remainders

Image


I understand answer has to be B.
I just want to understand question; As it is given P is a two digit integer written as 30a+b, where a and b is positive integer. Then option how can be the option A) a=4 is valid?

As per Gmat DS options are always true.

Please clarify. Thanks


Sent from my iPad using GMAT Club Forum mobile app


Hi,

Thanks for pointing it out.
Yes, you are right. "a" cannot be equal to 4. We have changed the statement now.

Regards,
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Re: P is a two-digit integer, which can be written as 30a + b, where a and  [#permalink]

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New post 30 Oct 2018, 06:21

Solution


Given:
In this question, we are given that,
    • P is a two-digit number.
    • P = 30a + b, where a and b are positive integers.

To find:
    • The remainder, when P is divided by 3.

Approach and Working:
We know that P can be expressed in the form 30a + b.
    • Hence, if P is divided by 3, we can actually divide 30a and b separately by 3, and then add the remainders to get the final answer.
    • However, irrespective of the value of a, 30a will be always divisible by 3.
      o Hence, if we get any remainder, that we will get when we will divide b by 3.

Therefore, we can conclude that, to determine the remainder when P is divided by 3, we need to know the value of b.

With this understanding, let’s now analyse the statements.

Analysing Statement 1
As per the information given in statement 1, a = 3.
    • However, this statement gives us no information about the value of b.

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

Analysing Statement 2
As per the information given in statement 2, \(b^3 – 5b^2 – 14b = 0\)

Simplifying the given equation, we get,
    • \(b (b^2 – 5b – 14) = 0\)
    Or,\(b (b^2 – 7b + 2b – 14) = 0\)
    Or, b [b (b – 7) + 2 (b – 7)] = 0
    Or, b (b – 7) (b + 2) = 0

Hence, b = -2 or 0 or 7

But we already know that b is positive.
    • Therefore, we can say b = 7.

As we can find the unique value of b from statement 2, we can conclude that statement 2 is sufficient to answer the question.

Combining Both Statements
Since we got an answer from the second statement individually, we don’t need to combine the statements.

Hence, the correct answer is option B.

Answer: B

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Re: P is a two-digit integer, which can be written as 30a + b, where a and   [#permalink] 30 Oct 2018, 06:21
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