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.

Answer: 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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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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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

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

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



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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

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.


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57 does not give the same remainder (3) when it is divided by 4,5, and 6 individually

My bad. I misinterpreted the Question.


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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 ..

Please kudos my effort/post if it helps ..

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.