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Five consecutive positive integers are chosen at random [#permalink]
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Updated on: 19 Sep 2013, 05:06
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Five consecutive positive integers are chosen at random. If the average of the five integers is odd, what is the remainder when the largest of the five integers is divided by 4? (1) The third of the five integers is a prime number. (2) The second of the five integers is the square of an integer.
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Originally posted by christykarunya on 19 Sep 2013, 03:22.
Last edited by Bunuel on 19 Sep 2013, 05:06, edited 1 time in total.
Added the OA.



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Re: Five consecutive positive integers are chosen at random [#permalink]
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19 Sep 2013, 03:41
stmt1: third of the five integers is a prime number. If the middle number is 7 then the largest number 9 which gives reminder 1 If the middle number is 5 then the largest number 7 which gives reminder 1
Insufficient
stmt2: the second of the integers is the square of an integer. so it must be divisible by 4 bcause the 2nd number is even number as per the question. so the largest number will be 4*2 + 3.
Sufficient.
Ans :B



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Re: Five consecutive positive integers are chosen at random [#permalink]
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20 Sep 2013, 03:04
neelzwiz wrote: stmt1: third of the five integers is a prime number. If the middle number is 7 then the largest number 9 which gives reminder 1 If the middle number is 5 then the largest number 7 which gives reminder 1
Insufficient
stmt2: the second of the integers is the square of an integer. so it must be divisible by 4 bcause the 2nd number is even number as per the question. so the largest number will be 4*2 + 3.
Sufficient.
Ans :B I think u did some typo hen explaining statement 1. If the middle number is 5 then the largest number 7 which gives reminder 3 Can u pls elaborate stmt 2 . I dint get that



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Re: Five consecutive positive integers are chosen at random [#permalink]
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20 Sep 2013, 06:49
christykarunya wrote: neelzwiz wrote: stmt1: third of the five integers is a prime number. If the middle number is 7 then the largest number 9 which gives reminder 1 If the middle number is 5 then the largest number 7 which gives reminder 1
Insufficient
stmt2: the second of the integers is the square of an integer. so it must be divisible by 4 bcause the 2nd number is even number as per the question. so the largest number will be 4*2 + 3.
Sufficient.
Ans :B I think u did some typo hen explaining statement 1. If the middle number is 5 then the largest number 7 which gives reminder 3 Can u pls elaborate stmt 2 . I dint get that Five consecutive positive integers are chosen at random. If the average of the five integers is odd, what is the remainder when the largest of the five integers is divided by 4?(1) The third of the five integers is a prime number. If the set is {1, 2, 3, 4, 5}, then the remainder of 5 divided by 4 is 1. If the set is {3, 4, 5, 6, 7), then the remainder of 7 divided by 4 is 3. Not sufficient. (2) The second of the five integers is the square of an integer. From the stem we know that the second integer must be even. An even number to be a square of an integer it must be a multiple of 4: 4, 16, 36, ... Thus, the largest integer is 4k+3, which means that the remainder is 3. Sufficient. Answer: B.
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Re: Five consecutive positive integers are chosen at random [#permalink]
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20 Sep 2013, 08:48
Bunuel wrote: christykarunya wrote: neelzwiz wrote: stmt1: third of the five integers is a prime number. If the middle number is 7 then the largest number 9 which gives reminder 1 If the middle number is 5 then the largest number 7 which gives reminder 1
Insufficient
stmt2: the second of the integers is the square of an integer. so it must be divisible by 4 bcause the 2nd number is even number as per the question. so the largest number will be 4*2 + 3.
Sufficient.
Ans :B I think u did some typo hen explaining statement 1. If the middle number is 5 then the largest number 7 which gives reminder 3 Can u pls elaborate stmt 2 . I dint get that Five consecutive positive integers are chosen at random. If the average of the five integers is odd, what is the remainder when the largest of the five integers is divided by 4?(1) The third of the five integers is a prime number. If the set is {1, 2, 3, 4, 5}, then the remainder of 5 divided by 4 is 1. If the set is {3, 4, 5, 6, 7), then the remainder of 7 divided by 4 is 3. Not sufficient. (2) The second of the five integers is the square of an integer. From the stem we know that the second integer must be even. An even number to be a square of an integer it must be a multiple of 4: 4, 16, 36, ... Thus, the largest integer is 4k+3, which means that the remainder is 3. Sufficient. Answer: B. Hi Bunuel could you please explain how we get 4k+3 here i am unable to understandho we get 3 there.



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Re: Five consecutive positive integers are chosen at random [#permalink]
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21 Sep 2013, 03:39
sivapavan wrote: Bunuel wrote: Five consecutive positive integers are chosen at random. If the average of the five integers is odd, what is the remainder when the largest of the five integers is divided by 4?
(1) The third of the five integers is a prime number.
If the set is {1, 2, 3, 4, 5}, then the remainder of 5 divided by 4 is 1. If the set is {3, 4, 5, 6, 7), then the remainder of 7 divided by 4 is 3.
Not sufficient.
(2) The second of the five integers is the square of an integer. From the stem we know that the second integer must be even. An even number to be a square of an integer it must be a multiple of 4: 4, 16, 36, ... Thus, the largest integer is 4k+3, which means that the remainder is 3. Sufficient.
Answer: B. Hi Bunuel could you please explain how we get 4k+3 here i am unable to understandho we get 3 there. The second of the five consecutive integers is 4k; The third of the five consecutive integers is 4k+1; The fourth of the five consecutive integers is 4k+2; The fifth of the five consecutive integers is 4k+3. Hope it's clear.
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Re: Five consecutive positive integers are chosen at random [#permalink]
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21 Sep 2013, 21:15
Hi. What am I missing to assume in the 2nd Statement that the 2nd integer could also be odd: 9, 25, 49, etc.?



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Re: Five consecutive positive integers are chosen at random [#permalink]
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22 Sep 2013, 01:23
reaching07 wrote: Hi. What am I missing to assume in the 2nd Statement that the 2nd integer could also be odd: 9, 25, 49, etc.? Let me try to explain... Assume numbers are x, x+1, x+2, x+3, x+4. Average = x+2= odd (given). so X+1 or second number will be even. So we can not assume 9,25,49 etc. Now why 4,16,36....? We know, Even * Even = Even Even * Odd = Even Odd * Odd = Odd But given number is Even and is square of a number (Ofcourse Even). So, Num= even * Even = 2k * 2k (each even can be written as 2k)= 4k^2 . putting k=1,2,3.... we get 4,16,36 ...etc. Hope it helps..
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Re: Five consecutive positive integers are chosen at random [#permalink]
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