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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

Answer C
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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

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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You are right, I missed it.
but N should be 60 not 63.

rahulkashyap
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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You are right, I missed it.
but N should be 60 not 63.

rahulkashyap
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


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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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
gmatbusters
You are right, I missed it.
but N should be 60 not 63.

rahulkashyap
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


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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57 does not give the same remainder (3) when it is divided by 4,5, and 6 individually
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My bad. I misinterpreted the Question. :cry:

rahulkashyap
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
Hence answer is D ..

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e-GMAT 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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What does the statement mean by "N is between 7 and 70"
is 7 and 70 inclusive?

if we consider 7 inclusive, then if we take n = 4, number of students = k. LCM (4, 5, 6) + 3 = 60k + 3 is valid.
but if we consider 7 exclusive then shouldn't we take n=5 which leads to number of students = k. LCM (5, 6, 7) + 3 = 210k + 3

And if that is true, then N is between 7 and 70 becomes invalid with N = 210k + 3

Kindly help

PS:- I am considering in n=5 in the 7 exclusive as N=5p + 3, if p = 1 then N = 8... Can anyone explain if my approach is wrong and how should i deal with such quant questions
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I thought hard questions were that way because they were hard or tricky to solve, not hard to read... If you understand what this question is asking, that's an achievement on its own. IMO this question is useless.
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­This is what I love about GMAT club. Various mentors/experts sharing multiple strategies to solve a questions. Tryuly amazing
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