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

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22 Jan 2017, 00:05

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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.

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

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22 Jan 2017, 05:40

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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.

Coming back to your question. 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.

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

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30 Jan 2017, 15:48

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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
_________________

I welcome analysis on my posts and kudo +1 if helpful. It helps me to improve my craft.Thank you

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

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30 Jan 2017, 15:58

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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
_________________

I welcome analysis on my posts and kudo +1 if helpful. It helps me to improve my craft.Thank you

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

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23 May 2017, 19:18

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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 [03], because 0 cannot play the role of tens digit.

Therefore, we have 14-1=13 numbers => (B) is correct

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.

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.

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.

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

My BAD.! 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

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.

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

My BAD.! 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 .....

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

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29 Dec 2017, 21:17

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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!

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

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30 Dec 2017, 05:11

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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!

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 !!
_________________

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

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30 Dec 2017, 06:15

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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

Althoughgmatexam439is 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

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