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e-GMAT Representative V
Joined: 04 Jan 2015
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If all the N students of a class are classified into groups of either  [#permalink]

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30 00:00

Difficulty:   95% (hard)

Question Stats: 12% (02:41) correct 88% (02:36) wrong based on 219 sessions

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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

To access all the questions: Question of the Week: Consolidated List

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Originally posted by EgmatQuantExpert on 07 Jun 2018, 22:31.
Last edited by EgmatQuantExpert on 12 Aug 2018, 23:52, edited 5 times in total.
e-GMAT Representative V
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1
2

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:
• What is the value of N

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 70

Hence, 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 number

The 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.

Important 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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##### General Discussion
Retired Moderator V
Joined: 27 Oct 2017
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Re: If all the N students of a class are classified into groups of either  [#permalink]

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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

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2
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.

Manager  B
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Re: If all the N students of a class are classified into groups of either  [#permalink]

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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

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
Retired Moderator V
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Re: If all the N students of a class are classified into groups of either  [#permalink]

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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.

Posted from my mobile device
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Re: If all the N students of a class are classified into groups of either  [#permalink]

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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.

Posted from my mobile device

N 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
Retired Moderator V
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Re: If all the N students of a class are classified into groups of either  [#permalink]

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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.

60-3 = 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.

Posted from my mobile device

N 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  [#permalink]

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57 does not give the same remainder (3) when it is divided by 4,5, and 6 individually
Retired Moderator V
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Re: If all the N students of a class are classified into groups of either  [#permalink]

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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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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 N-3 = L.C.M of { n ,(n+1) & (n+2) }

Let us check individual statements
Statement 1: N is B/W 7 and 70 ; or N-3 is B/W 4 & 67
minimum possible value of n can be 4 ..as inferred from the problem statement
With n=4 we have N-3= LCM of 4 ,5 & 6 ...=60 ..satisfies the constraint ...let us keep
Let us check for n=5 N-3=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

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  [#permalink]

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EgmatQuantExpert wrote:
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.

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