Why cant we choose option D here pls answer
n^2 is equal to the LCM or one of it’s multiples.
 LCM = 216
 n^2 = k.216.
 n = \sqrt{k.216} = 6\sqrt{k.6}
 Since n is an integer \sqrt{k.6} must be a perfect square, so k = 6 (at least), will give you n=36
 We want the options to divide all the items of set S. Since 36 is the least possible value in the set S, we can just test the options on it because the rest of the values will be anyways multiples of 36.
n^2 = \text{LCM}(24,108)
n=\sqrt{216\times k} , As 216 is not a perfect square
216 = 2^3 \times 3^3
n = \sqrt{2^3 \times 3^3 \times k(\text{where k =}2 \times 3)}
n = 36
Highest divisor of n = 36 is 36 and the only other option which divides n is A.12
Jinx
Look carefully they are asking for divisor of n and not multiple of n.
In this problem I do understand the concept that the number should at least have 2^2 and 3^2. The thing is, 72 equals 2^3 x 3^2, so it does meet the minimum requirement of 2^2 and 3^2, why is it not a correct choice?
If n = 2^2 \times3^2, then what is the highest divisor of n
I always get confused between terms like factors, divisors, multiples, etc. How can i learn those so that I remember??
Also i didnt understand this quest
a
isnt 72 a multiple of 36? it should be included right?
Let’s say A/B, gives ‘Q’ quotient and ‘R’ remainder
Divisor = b
If R = 0, then B and Q would be factors of A, and A would be a multiple of B and Q
Coming to the question, it says the S is the set of all positive integers n, such that n^2 is a multiple of 24 and 108
24 = 2^3*3
108 = 2^2*3^3
Smallest Multiple of 24 and 108 = 2^3 * 3^3
All other multiples of 24 and 108 have 2^3 * 3^3 multiplied by something
n^2 = x*2^3 * 3^3
n = y* 2^2 * 3^2
The question is asking for divisors of all possible values of n
12 = 2^2 * 3, can divide n
24 = 2^3 * 3 can divide n
36 = 2^2 * 3^2, can divide n
72 = 2^3 * 3^2, can divide some values of n not all
the question says “every” integer
In this question it says n^2 is a multiple of 108 AND 24. I factored 24 and 108 and found that n^2 must have min of 3 3’s and 3 2’s, and hence n must have atleast 2 3’s and 2 3’s, by this logic I eliminated A and B, but the answer key says B and C are correct. What did I do wrong?
108 = 2^2\times 3^3
24 = 2^3\times3
So, for n to be a multiple of both 108 & 24 it should have at minimum 2^2\times3^2 or 36. Now, divisor of 36 are 12 and 36.
you can’t square to the power of 3, thus you’ll have to do n^2=3^4\times2^4or,n=2^2\times3^2
or take the lcm = 216 = 2^3\times3^3 Now, we know that n^2 should have even power in its prime factor thus, we need to convert 2^3\times3^3 into even power .
Hello,
I’m super new to Gregmat and have just begun with the two month schedule. Had two queries:

There was a small homework question on ‘How many zeroes are there at the end of 25!?’
Query: Is it a correct approach to figure out the number 5’s in the prime factorization? Is the final answer two? 
In the things to do, under week 1 class 1, I didn’t completely understand the solution to the last problem  ‘S be the set of all positive integers n such that such that n^2 is a multiple of both 24 and 108…’
Query: Why did we consider only 2^3 for checking the divisibility of 24 and 3^2 for 108?
Thanks for your patience and time!
It is correct to figure out the no. of 5s in prime factorisation but the final answer would not be 2.
25! = 25x24x23 … 1
To know the no. of 5s, we look at all multiples of 5 up to 25:
5,10,15,20,25
The number of 5s that we get from each of these are:
1,1,1,1,2
Total no. of 5s = 1+1+1+1+2 =6
24 = 2³3
108 = 2²3³
If n² is a multiple of 24 and 108, hence it must be divisible by 2³3 and 2²3³
Because of case 1 we can say that n² is divisible by 2³ and 3
Because of case 2 we can say that n² is divisible by 2² and 3³
If we know that n² is divisible by 2³, then it will automatically be divisible by 2², and we don’t need to take the latter into consideration
Similarly for 3³ and 3
Hope this answers it