CAMANISHPARMAR wrote:
GyMrAT wrote:
3600 = 2^4*3^2*5^2
No. of factors of 3600 = 5 * 3 * 3 = 45
No. of factors of 3600 which are divisible by 6 are (45 - 11 -15 - 9 + 8 + 6 = 24) Correct ans is B
CAMANISHPARMARCould you please explain the methodology in obtaining this expression (45 - 11 -15 - 9 + 8 + 6 = 24), it looks very interesting.
Thanks,
GyM
GyMrATThank you for query, I am happy to help.
Logic:-
3600=2^4*3^2*6^2.
Number of factors = (4+1)(2+1)(2+1) = 45
The moment we have both a 2 and 3 in any factor then we will have a factor of 3600 which will be divisible by 6. Hence I removed all the combinations in which 2 & 3 both don't form a factor but in doing so I removed some factors twice hence I had to add them back. This was the logic. I have given you the hint in principle. You may try once by your own and still if you don't get it I will share the detailed calculations.
The aforesaid method is iterative and only for intellectual discussions
.....I recommend NOT to use this method in GMAT exam as you will be under time pressure. I have also posted a smarter way of tackling this question:-
https://gmatclub.com/forum/how-many-fac ... l#p2068935I hope this helps,
Regards,
Manish
Thanks Manish!!
I understand the principle that you are hinting at, below is the methodology i have understood. Please correct me if have gone wrong in my understanding. Its as you said for the sake of only intellectual discussion & not for practical use in GMAT. However its a good idea to strengthen the concept, definitely for me.
Ok, so i use combinations for identifying the different factors.
Now \(3600 = 2^4 * 3^2 * 5^2\)
The factors which contain only \(2\)'s can be combined as \(2^0, 2^1, 2^2, 2^3, 2^4\)
Similarly for \(3\), the factors will be \(3^0, 3^1, 3^2\)
& similarly for \(5\), the factors will be \(5^0, 5^1, 5^2\)
Total # of factors \(= 5*3*3 = 45\), which can derived with the formula as well
Now factors which contain \(6\) or in other words, divisible by \(6\), will have a combination that has \((2*3)\) as below
(\(2^1*3^1\)), (\(2^1*3^1*5^0\)), (\(2^1*3^2\)),(\(2^1*3^2*5^0\)), (\(2^1*3^1*5^1\)),...etc.
These are a combination of (\(2^1, 2^2, 2^3, 2^4\)), (\(3^1,3^2\)), (\(5^0, 5^1, 5^2\)), containing \((2*3)\)
So # of factors of \(3600\), divisible by \(6 = 4*2*3 = 24\)
Now i tried to find the factors not divisible by \(6\), just to tally with your expression \((45 - 11 -15 - 9 + 8 + 6 = 24)\). I could not found the exact numbers deducted or added by you, however i found the total factors not divisible by \(6\), using the same methodology used to find the factors divisible by 6. Please correct me, if i have gone wrong.
Factors which do not contain \((2*3)\), will include 2 scenarios, one in which we have all powers of \(2\) & \(3^0\) & other in which we consider all powers of \(3\) & only \(2^0\). Powers of \(5\) are included for both scenarios.
Case 1 - we have (\(2^0, 2^1, 2^2, 2^3, 2^4\)), \(3^0\), (\(5^0, 5^1, 5^2\)), hence \(5*1*3 = 15\) factors
Case 2 - we have \(2^0\), (\(3^0, 3^1,3^2\)), (\(5^0, 5^1, 5^2\)), hence \(1*3*3 = 9\) factors
Now we need to deduct the common factors from both cases. The common factors are (\(2^0*3^0*5^0\)), ((\(2^0*3^0*5^1\)) & (\(2^0*3^0*5^2\)), hence \(3\) factors.
So # of factors of \(3600\), not divisible by \(6 = 15+9-3 = 21\)
Let me know your thoughts on this.
Thanks,
GyM
15 - we have (\(2^0, 2^1, 2^2, 2^3, 2^4\)), \(3^0\), (\(5^0, 5^1, 5^2\)), hence \(5*1*3 = 15\) factors -
9 - we have \(2^0\), (\(3^0, 3^1,3^2\)), (\(5^0, 5^1, 5^2\)), hence \(1*3*3 = 9\) factors -
.....I recommend NOT to use this method in GMAT exam as you will be under time pressure. I have also posted a smarter way of tackling this question:-