How many factors of 3600 are divisible by 6?

Nice solution. I think you wanted to mention "5" here 3600=2^4*3^2*6^2..

3600 could be written as 6 * 600
Now 10 is a factor of 600, so is 20, so is 600 itself, including many others.
10 and 20 are not divisible by 6 but (6*10) and (6*20) are divisible by 6 because we are multiplying and dividing my 6. Hence every factor of 600 is also of factor of 3600 and are divisible by 6 if we are multiplying each factor of 600 by 6.
To find the number of factors of 600 is a straightforward application of number of factors formula:-
(p+1)(q+1)(r+1)... [where p,q,r are exponents of each prime factor]
Therefore 600 can be written as \(2^3*3*5^2\)
Therefore the number of factors of 600 are (3+1)(1+1)(2+1) = 24 factors.

Therefore the no. of factors of 3600 which are divisible by 6 is 24 factors.

Anybody knows where I can practice more of those specific questions? Thanks in advance.

Breaking down 3600 we see that:

3600 = 6 x 600 = 6 x (6 x 100) = 6 x (2 x 3 x 2^2 x 5^2) = 6 x (2^3 x 3^1 x 5^2)

We see that 600 = 2^3 x 3^1 x 5^2, which means 600 has (3+1) x (1+1) x (2+1) = 24 factors. Thus, we can pair 6 with any of the 24 factors of 600 to produce a product that is divisible by 6. Since each of these 24 pairings is a factor of 3,600, we have 24 factors of 3600 that are divisible by 6.

Answer: B

Hi,

Each question has tags in the top part of the screen. This Q has a tag: Divisibility/Multiples/Factors. Just click on it and enjoy.

CAMANISHPARMAR

Could you please explain the methodology in obtaining this expression (45 - 11 -15 - 9 + 8 + 6 = 24), it looks very interesting.


Thanks,
GyM

GyMrAT

Thank 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#p2068935

I 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

GyMrAT

Great try!!

Detailed calculations for (45−11−15−9+8+6=24) :-

Total # of factors \(= 5*3*3 = 45\)

11 - we have \(2^4\), (\(3^2\)), (\(5^2\)), hence \(5+3+3 = 11\) factors [We are individually considering all the factors of 2 first, then 3 and then 5, hence we are ADDING the individual No. of factors respectively]
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 - You had yourself got this number---good job!!
9 - we have \(2^0\), (\(3^0, 3^1,3^2\)), (\(5^0, 5^1, 5^2\)), hence \(1*3*3 = 9\) factors - You had yourself got this number---good job!!
8 - we have \(2^4\), (\(5^2\)), hence \(5+3 = 8\) factors - Adding back the no. of factors as we have subtracted them twice above.
6 - we have \(3^2\), (\(5^2\)), hence \(3+3 = 6\) factors - Adding back the no. of factors as we have subtracted them twice above.

I would like to reiterate:-

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#p2068935

All the best!! Once again GyMrAT - great try!!

Thanks Manish!! The concept is crystal clear now.

Yes on the real exam, its not feasible to use the detailed approach, however the concept clarity will be helpful, if the GMAT throws a curved ball on this topic.

Thanks again!!
GyM

GyMrAT

All the best my friend!! Nice meeting you and I am really impressed with your efforts.

Your curiosity to learn and use concepts has inspired and motivated me!!

Wishing you all the luck in this world which helps you get your dream score!!

Will keep in touch!! Let me know in case if I could be of any other help!!

I wish the same for you, Brother!! we are all together in this journey!!

I did a quasi guessing approach. I apologize ahead of time because my way of thinking was partially incorrect so if someone wants to point out my errors (which there are some), it would be greatly appreciated! That being said, I did get the correct answer:

What are the factors of 36?

1---------36
2---------18
3---------12
4----------9
6----------6
9 Total

Each factor can be multiplied by 1, 10, 100, 1000, you multiply 4*9 giving you 36 factors. We know that we are looking for factors divisible by 6, so we know that we can eliminate 9 (too few) and 45 (too many). It was at this exact moment that I went with the second highest answer.

:0 Apologies again haha but there might be a nugget of knowledge in there somewhere. That being said, I like the other solutions more.

Alternative:

3600 = 6*600

i.e. every factor of 600 when multiplied with 6 will become factor of 3600 which is also a multiple of 6

\(600 = 2^3*3^1*5^2\)

factors of \(1800 = (3+1)*(1+1)*(2+1) = 4*2*3 = 24\)

Answer: Option B

The video solution of the problem is attached here.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.