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If all the N students of a class are classified into groups of either
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Updated on: 12 Aug 2018, 23:52
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12% (02:41) correct 88% (02:36) wrong based on 219 sessions
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eGMAT Question of the Week #2If all the N students of a class are classified into groups of either n, or (n+1), or (n+2) students, every time 3 students are left out and cannot be included in any of those groups formed. What is the value of N? 1. N is between 7 and 70 2. n is the smallest number with exactly 3 distinct factors To access all the questions: Question of the Week: Consolidated List
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If all the N students of a class are classified into groups of either
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12 Jun 2018, 02:09
Solution Given:• If all the N students of a class are classified into groups of either n, or (n + 1), or (n + 2) students, every time 3 students are left out and cannot be included in any of those groups formed To find:Analysing Statement 1• As per the information provided in statement 1, N is between 7 and 70 As every time 3 students are left, the value of n is greater than 3 • If we take n = 4, number of students = k. LCM (4, 5, 6) + 3 = 60k + 3
o As N is between 7 and 70, the only possible value is 63, when k = 1 For any other values of n, the value of N does not lie in between 7 and 70Hence, statement 1 is sufficient to answer Analysing Statement 2• As per the information provided in statement 2, n us the smallest number with exactly 3 distinct factors When a number has exactly 3 distinct factors, it is the square of a prime numberThe smallest of such numbers = 4 Therefore, we can say the values of N will be in the form k. LCM (4, 5, 6) + 3 or 60k + 3 • If k = 1, N = 63 • If k = 2, N = 123 • If k = 3, N = 183 and so on As we don’t know the exact value of k, we can’t find out the exact value of N Hence, statement 2 is not sufficient to answer Hence, the correct answer is option A. Answer: AImportant Observation This question is an example where one can easily conclude that statement 1 is not sufficient or statement 2 is sufficient to answer, without observing the inferences present in the statements. • As per statement 1, in the given range only one value is possible. One needs to check whether any other possibilities exist or not  if not then this statement is giving us a unique answer, and hence, sufficient to answer.
• As per statement 2, although one can conclude n = 4, it only helps us in obtaining the general form of N. the number 63 is only one of the possible cases and not necessarily the only case.
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Re: If all the N students of a class are classified into groups of either
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07 Jun 2018, 22:53
If all the N students of a class are classified into groups of either n, or (n+1), or (n+2) students, every time 3 students are left out and cannot be included in any of those groups formed. What is the value of N? Since all the N students of a class are classified into groups of either n, or (n+1), or (n+2) students, it means N = Multiple of LCM of n, (n+1), & (n+2). since we do not know n, LCM cannot be found, only we know that LCM is multiple of 3! or 6 1. N is between 7 and 70. since n is not known, N cannot be found. INSUFFICIENT2. n is the smallest number with exactly 3 distinct factors n = 2^2 = 4, So N is multiple of LCM of 4,5,6 or multiple of 60. N can be 60, (57 is not divisible by 4,5,6) or 120 (57 is not divisible by 4,5,6). Value of N cannot be uniquely determined, INSUFFICIENTCombining Statement 1 & 2, we get N = multiple of LCM of 60 and N is between 7 and 70. Hence N = 60. SUFFICIENTAnswer C
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Re: If all the N students of a class are classified into groups of either
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07 Jun 2018, 23:21
assuming this question means that the remainder when N is divided by n, n+1 or n+2 is 3, we can calculate N by finding out the LCM of these 3 numbers.
First of all n has to be greater than 3, as the remainder itself is 3.
First set 4,5,6 LCM = 60. N= 60+3= 63. Second set  5,6,7 LCM= 210 . N = 210+3. and so on
As A says N is between 7 and 70, only one set, i.e. 4 , 5 and 6 is suitable for this, hence the answer is A.
B says n( the first number) is a perfect square, i.e. n is 4,9,25,36, and so on. Multiple sets can be formed.
A should be the answer



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Re: If all the N students of a class are classified into groups of either
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07 Jun 2018, 23:25
gmatbusters wrote: If all the N students of a class are classified into groups of either n, or (n+1), or (n+2) students, every time 3 students are left out and cannot be included in any of those groups formed. What is the value of N?
Since all the N students of a class are classified into groups of either n, or (n+1), or (n+2) students, it means N = Multiple of LCM of n, (n+1), & (n+2). since we do not know n, LCM cannot be found, only we know that LCM is multiple of 3! or 6
1. N is between 7 and 70. since n is not known, N cannot be found. INSUFFICIENT
2. n is the smallest number with exactly 3 distinct factors n = 2^2 = 4, So N is multiple of LCM of 4,5,6 or multiple of 60. N can be 60, (57 is not divisible by 4,5,6) or 120 (57 is not divisible by 4,5,6). Value of N cannot be uniquely determined, INSUFFICIENT
Combining Statement 1 & 2, we get N = multiple of LCM of 60 and N is between 7 and 70.
Hence N = 60. SUFFICIENT
Answer C for a) take examples. 4,5,6 or 5,6,7. You will see that only one set satisfies this requirement, i.e. 4,5,6



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Re: If all the N students of a class are classified into groups of either
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08 Jun 2018, 02:44
You are right, I missed it. but N should be 60 not 63. rahulkashyap wrote: assuming this question means that the remainder when N is divided by n, n+1 or n+2 is 3, we can calculate N by finding out the LCM of these 3 numbers.
First of all n has to be greater than 3, as the remainder itself is 3.
First set 4,5,6 LCM = 60. N= 60+3= 63. Second set  5,6,7 LCM= 210 . N = 210+3. and so on
As A says N is between 7 and 70, only one set, i.e. 4 , 5 and 6 is suitable for this, hence the answer is A.
B says n( the first number) is a perfect square, i.e. n is 4,9,25,36, and so on. Multiple sets can be formed.
A should be the answer Posted from my mobile device
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Re: If all the N students of a class are classified into groups of either
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08 Jun 2018, 02:59
gmatbusters wrote: You are right, I missed it. but N should be 60 not 63. rahulkashyap wrote: assuming this question means that the remainder when N is divided by n, n+1 or n+2 is 3, we can calculate N by finding out the LCM of these 3 numbers.
First of all n has to be greater than 3, as the remainder itself is 3.
First set 4,5,6 LCM = 60. N= 60+3= 63. Second set  5,6,7 LCM= 210 . N = 210+3. and so on
As A says N is between 7 and 70, only one set, i.e. 4 , 5 and 6 is suitable for this, hence the answer is A.
B says n( the first number) is a perfect square, i.e. n is 4,9,25,36, and so on. Multiple sets can be formed.
A should be the answer Posted from my mobile deviceN has to be 63 and not 60, as 60 divided by 4,5 and 6 gives a remainder of 0. 63 gives a remainder of 3 when divided by all the 3 numbers, and that is what we need



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Re: If all the N students of a class are classified into groups of either
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08 Jun 2018, 04:19
yes it has to be multiple of 4, 5 and 6. Question is that if 3 are removed, then it should not be divisible by n, n+1, n+2. 603 = 57 is not divisible by 4,5 or 6. rahulkashyap wrote: gmatbusters wrote: You are right, I missed it. but N should be 60 not 63. rahulkashyap wrote: assuming this question means that the remainder when N is divided by n, n+1 or n+2 is 3, we can calculate N by finding out the LCM of these 3 numbers.
First of all n has to be greater than 3, as the remainder itself is 3.
First set 4,5,6 LCM = 60. N= 60+3= 63. Second set  5,6,7 LCM= 210 . N = 210+3. and so on
As A says N is between 7 and 70, only one set, i.e. 4 , 5 and 6 is suitable for this, hence the answer is A.
B says n( the first number) is a perfect square, i.e. n is 4,9,25,36, and so on. Multiple sets can be formed.
A should be the answer Posted from my mobile deviceN has to be 63 and not 60, as 60 divided by 4,5 and 6 gives a remainder of 0. 63 gives a remainder of 3 when divided by all the 3 numbers, and that is what we need
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Re: If all the N students of a class are classified into groups of either
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08 Jun 2018, 04:28
57 does not give the same remainder (3) when it is divided by 4,5, and 6 individually



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Re: If all the N students of a class are classified into groups of either
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08 Jun 2018, 05:02
My bad. I misinterpreted the Question. rahulkashyap wrote: 57 does not give the same remainder (3) when it is divided by 4,5, and 6 individually
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Re: If all the N students of a class are classified into groups of either
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08 Jun 2018, 09:55
Given : Total students are N , if we divide them in three groups of n ,(n+1) & (n+2) ..every time 3 students are left or remainder is 3 and it can not be included in any group ..that means n>3 .
Approach : N=L.C.M of{ n ,(n+1) & (n+2) }+3
or N3 = L.C.M of { n ,(n+1) & (n+2) }
Let us check individual statements Statement 1: N is B/W 7 and 70 ; or N3 is B/W 4 & 67 minimum possible value of n can be 4 ..as inferred from the problem statement With n=4 we have N3= LCM of 4 ,5 & 6 ...=60 ..satisfies the constraint ...let us keep Let us check for n=5 N3=LCM of 5,6 & 7 ...=210 ...does not satisfy the constraint ... We can say only possible value of n is 4 ...and hence N is 63...statement 1 is sufficient ..
Let us see statement 2 ..which gives additional information of n is the smallest number with exactly three distinct factors ..inferring n must be a positive integer as number of students can be a positive integer only .. n=P1^a*P2^b.. ..number of distinct factors are (a+1)(b+1)=1*3 a+1=1 ; and b+1=3 a=0 and b=2...power is two and the smallest prime number is 2 ...hence n must be 2^2=4....hence statement 2 alone is sufficient .. N is = LCM of 4 , 5 & 6 +3 =63 Hence answer is D ..
Please kudos my effort/post if it helps ..



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Re: If all the N students of a class are classified into groups of either
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08 Jun 2018, 10:23
EgmatQuantExpert wrote: eGMAT Question of the Week #2 If all the N students of a class are classified into groups of either n, or (n+1), or (n+2) students, every time 3 students are left out and cannot be included in any of those groups formed. What is the value of N? 1. N is between 7 and 70 2. n is the smallest number with exactly 3 distinct factors Given N = nk + 3 N = (n+1)p + 3 N = (n+2)q + 3 Hence N = LCM of (n, (n+1), (n+2)) * x + 3 Also n>3, lets say n =4, then LCM of (4,5,6) is 60, hence N = 60x + 3 = 63, 123, 183,...etc. lets say n = 5, then LCM of (5,6,7) = 210, hence N = 210x +3 = 213, 423,...etc, similarly for other consecutive numbers. Statement 1: 7<N<70 From our upfront work, we can see that N = 60(1) + 3 = 63, is the only # that satisfies constraint of statement 1. Hence Statement 1 alone is Sufficient. Statement 2: n is the smallest number with three distinct factors. Hence n is a square of a prime number & since it needs to be the smallest, \(n = 2^2 = 4\) So from our upfront work, we can say N = 60x + 3 = 63, 123, 183,...etc. Statement 2 gives multiple values of N. Hence Statement 2 alone is Insufficient. Answer A.
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Re: If all the N students of a class are classified into groups of either
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29 Jun 2019, 06:33
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