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GRE 1: Q169 V154 Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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Mock Test 3 (600-700) Updated.

http://gmatclub.com/forum/stonecold-s-mock-test-217160.html#p1768023
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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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Hi Stonecold ! First of all thanks for sharing these awesome conceptual questions. I have one doubt - Why "Z" cannot be -ve (Negative integer) for question no 1 in Mock test1

1)If z is divisible by all the integers from 1 to 5 inclusive,what is the smallest value of z?
Possible minimum values for Z also include -60,-120 etc. 60 should be the smallest positive integer.

Please correct me if i am wrong.
Current Student D
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GRE 1: Q169 V154 STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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Vianand4 wrote:
Hi Stonecold ! First of all thanks for sharing these awesome conceptual questions. I have one doubt - Why "Z" cannot be -ve (Negative integer) for question no 1 in Mock test1

1)If z is divisible by all the integers from 1 to 5 inclusive,what is the smallest value of z?
Possible minimum values for Z also include -60,-120 etc. 60 should be the smallest positive integer.

Please correct me if i am wrong.

Hi.
Thanks for the feedback on the Quiz.

Technically you are right.

But the Bottom line is -> Factors and multiples on the GMAT are always positive integers.
Hence z would always be 60.

Regards
Stone Cold
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GRE 1: Q169 V154 Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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Mock Test 4 (600-700) Updated.

http://gmatclub.com/forum/stonecold-s-mock-test-217160.html#p1768052
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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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1)If z is divisible by all the integers from 1 to 5 inclusive,what is the smallest value of z?

Without any options how can we think and try z = 60?
Yes I understand 60 is divisible by all numbers 1 to 5
Assume If z is divisible by all integers from 1 to 7 inclusive, what is the smallest value of z? without knowing any options?
Thanks
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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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2)Data Sufficiency-> Is the positive integer z divisible by 12?
A)z is divisible by 3.
B)z is not divisible by 2.

1, z / 3 = integer
z could be 3,6,9,12,15,24...
Insufficient
BCE
2, z / 2 # integer
z must be an odd number
Sufficient

3)Data Sufficiency-> Is integer z divisible by 21?
A)z is divisible by 7 --- Insufficient
B)z is divisible by 3 ---- Insufficient
Z = 21, divisible by both 3 and 7
C

4, E
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GRE 1: Q169 V154 STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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A quadrilateral has a perimeter of 16. Which of the following alone would provide sufficient information to determine the area of the quadrilateral.

B)The quadrilateral is formed by combining two isosceles right triangles.

C) Two pairs of congruent angles are in a 2:1 ratio.

D) The width is 4o% of the length and all angles are of equal measure.

E) If the perimeter was decreased by 50%, the area would decrease by 25%

D.

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GRE 1: Q169 V154 Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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The difference between the digits of a two digit number is 3.How many such numbers are possible ?
A)12
B)13
C)14
D)15
E)16

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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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stonecold wrote:
The difference between the digits of a two digit number is 3.How many such numbers are possible ?
A)12
B)13
C)14
D)15
E)16

Hello, I am new in your topic. Is there anyone still discussing this?

The following is my take. Hope to receive your comment on it. Thank you.

Difference between 2 digits is 3 => Smaller digit must range from 0 to 6 (because if the smaller is 7, then the larger will be 10 -> this cannot happen).
The difference is constant, so we have 7 numbers
Tens digit will be either the larger or smaller, so we have 7x2=14 number
Eliminate the case , because 0 cannot play the role of tens digit.

Therefore, we have 14-1=13 numbers => (B) is correct
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If 2,3,5 are the only prime factors of a positive integer p,what is th  [#permalink]

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Sorry I am unable to find the existing thread for this question:

If 2,3,5 are the only prime factors of a positive integer p,what is the value of p?
A)p>100
B)p has exactly 12 factors including 1 and 12.

Spoiler: :: OA
C
.
Spoiler: :: Explanation
Since 2,3,5 are the only prime factors of p=> The general expression for p will be 2a∗3b∗5c2a∗3b∗5c where a,b,c are positive integers.
Statement 1->p>100.
There exist infinite such cases.
In-fact there are only a few cases where p will be less than 100.
Hence not sufficient.
Statement 2->
Factors =12
Hence (a+1)*(b+1)*(c+1)=12=2*2*3
NOTE-> We can't arrange 12 as 1*1*12 or 2*6*1 or 4*3*1 as a,b,c are positive integers.
Hence a,b,c => (1,1,2) or (1,2,1) or (2,1,1)
Posible values of p=> 150,90,60
Hence not sufficient.
Combining the two statements => p must be 150.
Hence C.

B

Spoiler: :: My understanding
The option B explicitly states that 12 is a factor of p. So, 2 must have a power of 2 and 3 must have power 1. Now since number of factors is 12, 5 should have power=1.
Thus only feasible solution is p=60.

Please point out the issue in my understanding

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GMAT 1: 630 Q47 V29 GMAT 2: 740 Q51 V38 Re: If 2,3,5 are the only prime factors of a positive integer p,what is th  [#permalink]

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12 is not a factor of 150 or 90...so U know the answer

Sent from my iPhone using GMAT Club Forum
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Re: If 2,3,5 are the only prime factors of a positive integer p,what is th  [#permalink]

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Luckisnoexcuse wrote:
12 is not a factor of 150 or 90...so U know the answer

Sent from my iPhone using GMAT Club Forum

Hi bro .. how is prep?

Please have a look at the official explanation. i think that is wrong, hence my doubt
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GRE 1: Q169 V154 Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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gmatexam439 wrote:
Sorry I am unable to find the existing thread for this question:

If 2,3,5 are the only prime factors of a positive integer p,what is the value of p?
A)p>100
B)p has exactly 12 factors including 1 and 12.

Spoiler: :: OA
C
.
Spoiler: :: Explanation
Since 2,3,5 are the only prime factors of p=> The general expression for p will be 2a∗3b∗5c2a∗3b∗5c where a,b,c are positive integers.
Statement 1->p>100.
There exist infinite such cases.
In-fact there are only a few cases where p will be less than 100.
Hence not sufficient.
Statement 2->
Factors =12
Hence (a+1)*(b+1)*(c+1)=12=2*2*3
NOTE-> We can't arrange 12 as 1*1*12 or 2*6*1 or 4*3*1 as a,b,c are positive integers.
Hence a,b,c => (1,1,2) or (1,2,1) or (2,1,1)
Posible values of p=> 150,90,60
Hence not sufficient.
Combining the two statements => p must be 150.
Hence C.

B

Spoiler: :: My understanding
The option B explicitly states that 12 is a factor of p. So, 2 must have a power of 2 and 3 must have power 1. Now since number of factors is 12, 5 should have power=1.
Thus only feasible solution is p=60.

Please point out the issue in my understanding

You are absolutely correct.
The OA should definitely be B.

if 12 is the factor the only possible combination would be 2^2*3*5

Best
Stone

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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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stonecold wrote:
gmatexam439 wrote:
Sorry I am unable to find the existing thread for this question:

If 2,3,5 are the only prime factors of a positive integer p,what is the value of p?
A)p>100
B)p has exactly 12 factors including 1 and 12.

Spoiler: :: OA
C
.
Spoiler: :: Explanation
Since 2,3,5 are the only prime factors of p=> The general expression for p will be 2a∗3b∗5c2a∗3b∗5c where a,b,c are positive integers.
Statement 1->p>100.
There exist infinite such cases.
In-fact there are only a few cases where p will be less than 100.
Hence not sufficient.
Statement 2->
Factors =12
Hence (a+1)*(b+1)*(c+1)=12=2*2*3
NOTE-> We can't arrange 12 as 1*1*12 or 2*6*1 or 4*3*1 as a,b,c are positive integers.
Hence a,b,c => (1,1,2) or (1,2,1) or (2,1,1)
Posible values of p=> 150,90,60
Hence not sufficient.
Combining the two statements => p must be 150.
Hence C.

B

Spoiler: :: My understanding
The option B explicitly states that 12 is a factor of p. So, 2 must have a power of 2 and 3 must have power 1. Now since number of factors is 12, 5 should have power=1.
Thus only feasible solution is p=60.

Please point out the issue in my understanding

You are absolutely correct.
The OA should definitely be B.

if 12 is the factor the only possible combination would be 2^2*3*5

Best
Stone

hehe finally i succeeded in finding one wrong ..... Regards
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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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Hi,

Thank you for posting awesome test. I have a question though. For Mock 3 question # 2, I am not able to follow the second stem and wondering if you could help me clarify my lack of understanding.

2)Data Sufficiency->If n is a positive integer,is n divisible by 2?
A)7n-8 is divisible by 20.
B)3n^2+2n+5 is a prime number.

I get that A is sufficient but I can't figure out B. Answer is D. Can you please stonecold help me figure this out. Thank you!
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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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Bobzi wrote:
Hi,

Thank you for posting awesome test. I have a question though. For Mock 3 question # 2, I am not able to follow the second stem and wondering if you could help me clarify my lack of understanding.

2)Data Sufficiency->If n is a positive integer,is n divisible by 2?
A)7n-8 is divisible by 20.
B)3n^2+2n+5 is a prime number.

I get that A is sufficient but I can't figure out B. Answer is D. Can you please stonecold help me figure this out. Thank you!

Hi Bobzi,

Statement 2:
$$3n^2+2n+5$$ is a prime number. Now since n is a positive integer, it means that n >=1. If that is the case then, $$3n^2+2n+5$$ would always be greater than 2. So for $$3n^2+2n+5$$ to be prime, it ought to be odd.

Start with n=1: $$3n^2+2n+5$$=10 --Even; Thus $$3n^2+2n+5$$ can never be prime when n=odd
With, n=2: $$3n^2+2n+5$$=21 --Odd; Thus $$3n^2+2n+5$$ will always be odd when n=even and thus there will exist a number which will be prime [we need not find that number].

So we can conclude that n=even and even numbers are always divisible by 2. --Sufficient

I hope that clears your doubt !! _________________
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GRE 1: Q169 V154 Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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Bobzi wrote:
Hi,

Thank you for posting awesome test. I have a question though. For Mock 3 question # 2, I am not able to follow the second stem and wondering if you could help me clarify my lack of understanding.

2)Data Sufficiency->If n is a positive integer,is n divisible by 2?
A)7n-8 is divisible by 20.
B)3n^2+2n+5 is a prime number.

I get that A is sufficient but I can't figure out B. Answer is D. Can you please stonecold help me figure this out. Thank you!

Hi Bobzi

Although
gmatexam439 is spot on with his method, I would like to use a more fundamental approach for statement 2.

We need to use the basics.
Remember all primes >2 are odd? Yes, we will use that here.

We are told 3n^2+2n+5 is a prime number
Since n is a positive integer the least possible value of n is one.

So the least possible value of 3n^2+2n+5 is 3+3+5 => 11
Hence 3n^2+2n+5 is a prime number ≥11

What is the common thing about all primes greater than 2? They are odd.

Hence 3n^2+2n+5 is odd.

So 3n^2+ even + odd = odd
3n^2 + odd = odd
3n^2 = odd-odd = even

3n^2 = even

Hence n = even.

Hence sufficient
Smash that D.

Best
Stone

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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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Excellent mate , Can we have answer explanations for these ? To be honest I'm getting many errors , would like to have some explanations . If you could help ?
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GRE 1: Q169 V154 Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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loserunderachiever wrote:
Excellent mate , Can we have answer explanations for these ? To be honest I'm getting many errors , would like to have some explanations . If you could help ?

Hi,
Which questions do you want me to discuss?

I am happy to help with any quant questions you have for me.
Just post in your question below and I will post in the solution.

Best
Stone
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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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Stone ,

Can you give me a solution of these two ?
Attachments Screen Shot 2018-03-03 at 12.57.18 PM.png [ 104.42 KiB | Viewed 1738 times ]

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" Can't stop learning and failing" Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.   [#permalink] 03 Mar 2018, 00:26

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