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M12-21

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M12-21  [#permalink]

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New post 16 Sep 2014, 00:47
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A
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Re M12-21  [#permalink]

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New post 16 Sep 2014, 00:47
1
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Official Solution:


How many integers in a group of 5 consecutive positive integers are divisible by 4?

(1) The median of these numbers is odd. The median of the set with odd number of terms is just a middle term, thus our set of 5 consecutive numbers is: {Odd, Even, Odd, Even, Odd}. Out of 2 consecutive even integers only one is a multiple of 4. Sufficient.

(2) The average (arithmetic mean) of these numbers is a prime number. In any evenly spaced set the arithmetic mean (average) is equal to the median, so we have that mean=median=prime. Since it's not possible that median=2=even, (in this case not all 5 numbers will be positive), then median=odd prime, and we have the same case as above. Sufficient.


Answer: D
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New post 05 Jan 2016, 14:07
I understood solution, but I just wanted to know why exactly does 1 of the number will be divisible by 4? Is their any kind of rule here that i have missed. I took various different sets of positive consecutive numbers with median being odd and in all cases,one of the number was divisible by 4.
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New post 06 Jan 2016, 08:13
rhio wrote:
I understood solution, but I just wanted to know why exactly does 1 of the number will be divisible by 4? Is their any kind of rule here that i have missed. I took various different sets of positive consecutive numbers with median being odd and in all cases,one of the number was divisible by 4.


I think this is explained in the solution: out of 2 consecutive even integers only one is a multiple of 4.
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Re: M12-21  [#permalink]

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New post 12 Jan 2016, 05:40
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rhio wrote:
I understood solution, but I just wanted to know why exactly does 1 of the number will be divisible by 4? Is their any kind of rule here that i have missed. I took various different sets of positive consecutive numbers with median being odd and in all cases,one of the number was divisible by 4.


because the other no. will yield the reminder 2 and hence not divisible
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Re: M12-21  [#permalink]

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New post 12 Jan 2016, 07:48
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rhio wrote:
I understood solution, but I just wanted to know why exactly does 1 of the number will be divisible by 4? Is their any kind of rule here that i have missed. I took various different sets of positive consecutive numbers with median being odd and in all cases,one of the number was divisible by 4.



Hi,
In all cases, any number, n, will have only one number divisible by itself in 'n' consecutive numbers..
5 will have only one multiple in any 5 consecutive number because of the property of multiplication tables..
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Re: M12-21  [#permalink]

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New post 15 Jul 2016, 22:57
can someone explain me me how option 2 is right?

for eg 7 8 9 10 11
9 is a median and not prime so cannot determine . Ans should be just option A
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New post 15 Jul 2016, 23:13
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Re: M12-21  [#permalink]

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New post 16 Jul 2016, 01:29
I considered 2 cases 1 2 3 4 5 and 7 8 9 10 11, for one we can determine and for other cannot .I misunderstood the question. Thanks for the help
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Re M12-21  [#permalink]

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New post 18 Nov 2017, 07:27
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Is 2 divisible by 4?

Ans would be 1/2. Does it satisfy our condition?
Shouldn't we look for a whole number as an answer?

In refrence Statement 1, considering the set : (1, 2, 3, 4, 5)
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Re: M12-21  [#permalink]

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New post 18 Nov 2017, 07:47
jasanisanket24 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Is 2 divisible by 4?

Ans would be 1/2. Does it satisfy our condition?
Shouldn't we look for a whole number as an answer?

In refrence Statement 1, considering the set : (1, 2, 3, 4, 5)


Hi jasanisanket24

the question asks how many numbers in the set are divisible by 4. so 2 is not divisible by 4. Hence in the set (1,2,3,4,5) we have only 1 number that is divisible by 4. Hence the statement is sufficient.

Both statement 1 & 2 leads us to the conclusion that the consecutive set starts with an ODD integer. so the set will be of the form (odd,even,odd,even,odd)

hence there will be two even integers in the set and both will be consecutive even integers. out of any two consecutive even integers, only 1 is divisible by 4 (eg. 2,4; 4,6; 8,10....etc). SO we can conclude that there is only 1 number that is divisible by 4. Hence both the statements are sufficient
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Re: M12-21  [#permalink]

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New post 25 Nov 2019, 23:45
Bunuel wrote:
Official Solution:


How many integers in a group of 5 consecutive positive integers are divisible by 4?

(1) The median of these numbers is odd. The median of the set with odd number of terms is just a middle term, thus our set of 5 consecutive numbers is: {Odd, Even, Odd, Even, Odd}. Out of 2 consecutive even integers only one is a multiple of 4. Sufficient.

(2) The average (arithmetic mean) of these numbers is a prime number. In any evenly spaced set the arithmetic mean (average) is equal to the median, so we have that mean=median=prime. Since it's not possible that median=2=even, (in this case not all 5 numbers will be positive), then median=odd prime, and we have the same case as above. Sufficient.


Answer: D


For St1 What about -1,0,1,2,3; Median is odd but none is divisible by 4
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New post 25 Nov 2019, 23:48
Nityanshu1990 wrote:
Bunuel wrote:
Official Solution:


How many integers in a group of 5 consecutive positive integers are divisible by 4?

(1) The median of these numbers is odd. The median of the set with odd number of terms is just a middle term, thus our set of 5 consecutive numbers is: {Odd, Even, Odd, Even, Odd}. Out of 2 consecutive even integers only one is a multiple of 4. Sufficient.

(2) The average (arithmetic mean) of these numbers is a prime number. In any evenly spaced set the arithmetic mean (average) is equal to the median, so we have that mean=median=prime. Since it's not possible that median=2=even, (in this case not all 5 numbers will be positive), then median=odd prime, and we have the same case as above. Sufficient.


Answer: D


For St1 What about -1,0,1,2,3; Median is odd but none is divisible by 4


0 is divisible by every integer except 0 itself: 0/integer = integer.
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New post 26 Nov 2019, 08:09
This is an excellent question on statistics and number properties. From the question statement, we know that the integers are consecutive and that they are all positive integers.

In a group of n consecutive integers, one integer will always be a multiple of n. For example, in a set of 3 consecutive integers, there will always be a multiple of 3. This is the underlying concept that we can use to solve this question.

However, we need to be careful about the possibility that there could be two multiples of 4 in a set of 5 consecutive integers. For example, 4,5,6,7,8 contains two integers which are divisible by 4. This is where we use the statements to evaluate how many of them can be divisible by 4.

Let us represent the 5 consecutive integers as a, (a+1), (a+2), (a+3) and (a+4). Remember that we have already taken these in order since these are consecutive.

From statement I, the median of these numbers is odd. This means that (a+2) is odd, which means that there are 3 odd numbers and 2 even numbers. In this case, only one of the even numbers will be divisible by 4. A set of 5 numbers will have 2 multiples of 4 when the first number itself is a multiple of 4, because of the rule stated above.. Clearly, that’s not happening here.

Statement I is sufficient to say that the set of integers consists of ONE number that is divisible by 4. Possible answer options are A or D. Answer options B, C and E can be eliminated.

From statement II, the average of the given numbers is a prime number.


For equally spaced values in a data set, Mean = Median.
Consecutive integers definitely represent equally spaced values. So, for the numbers that we have considered, the average(mean) is going to be the middle value.

So, the average is (a+2).
If(a+2) = 2, a = 0 which is not possible because all the numbers in the set are positive integers.

Therefore, (a+2) has to be an odd prime number. This is equivalent to the data given in the first statement. Since statement I alone was sufficient, statement II alone will also be sufficient. This is a very simple and logical conclusion.

Answer option A can be eliminated. The correct answer option is D.

Hope that helps!
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Re: M12-21  [#permalink]

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New post 26 Nov 2019, 08:14
Nityanshu1990 wrote:
Bunuel wrote:
Official Solution:


How many integers in a group of 5 consecutive positive integers are divisible by 4?

(1) The median of these numbers is odd. The median of the set with odd number of terms is just a middle term, thus our set of 5 consecutive numbers is: {Odd, Even, Odd, Even, Odd}. Out of 2 consecutive even integers only one is a multiple of 4. Sufficient.

(2) The average (arithmetic mean) of these numbers is a prime number. In any evenly spaced set the arithmetic mean (average) is equal to the median, so we have that mean=median=prime. Since it's not possible that median=2=even, (in this case not all 5 numbers will be positive), then median=odd prime, and we have the same case as above. Sufficient.


Answer: D


For St1 What about -1,0,1,2,3; Median is odd but none is divisible by 4


I think the question says that we have 5 consecutive positive integers. -1 is not a positive integer. Please be careful, do not miss out on such pieces of information given in the question. When you try out a case like this, please re-read the question just to ensure that you are not missing out on information.

Thanks
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Re: M12-21  [#permalink]

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New post 28 Nov 2019, 01:43
Bunuel wrote:
Official Solution:


(2) The average (arithmetic mean) of these numbers is a prime number. In any evenly spaced set the arithmetic mean (average) is equal to the median, so we have that mean=median=prime. Since it's not possible that median=2=even, (in this case not all 5 numbers will be positive), then median=odd prime, and we have the same case as above. Sufficient.



So '0' is considered neither negative nor positive for all purposes on the GMAT?

Bcz [0,1,2,3,4] has a median=even=2 and contradicts St(2).
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New post 28 Nov 2019, 04:07
RaunaqSinghPunn wrote:
Bunuel wrote:
Official Solution:


(2) The average (arithmetic mean) of these numbers is a prime number. In any evenly spaced set the arithmetic mean (average) is equal to the median, so we have that mean=median=prime. Since it's not possible that median=2=even, (in this case not all 5 numbers will be positive), then median=odd prime, and we have the same case as above. Sufficient.



So '0' is considered neither negative nor positive for all purposes on the GMAT?

Bcz [0,1,2,3,4] has a median=even=2 and contradicts St(2).


0 is neither positive nor negative generally in math, not only for the GMAT.
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