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I am quite close to my aim so I don't feel that I can justify buying your full pack, but your free stuff is amazing and whether or not I end up investing in the full quant I just want to say thank you so much for everything you have made available. It has been very useful and I'd easily recommend you to anyone I know.
I got the ans as 15 and let me explain before I see your video for answer. -I eliminated top 2 as none of the factorials are greater than 1000 -in 9 factorial, I searched for 3 unique combos whose multiplication results in multiples of 10. Only 5x4...hence ruled out - In 15 factorial, I searched 3 unique combinations whose multiplications result is a muliple of 10 i.e. 1×10, 5x6, 15x2 .... This means that the whole result set of 15! is divisible by (10x10x10) i.e. 1000 Hence, 15 is the answer
Thank you for the kind words. I'm glad it motivated you... and hopefully helped you better understand the principles to get similar questions right on test day!
thank you! appreciate it! I'm currently facing a difficult period as my new job start date has been postponed few time, and I found your videos very helpful. This is the second one I'm watching.
Awesome, Chloel, I'm glad to hear it and thanks for sharing. I'm confident you'll find the other videos on this channel helpful and then whenever you're ready to dive even deeper, I hope you'll consider one of our paid courses. Best of luck and let me know how else I can help!
Hey Brett. I was able to eliminiate answer choices A - C from the off. Just a question. Could we break down choice D / E as follows. 15! = (3 x 5)! Or 30! as (2 x 3 x 5)! and then solve.
Use Legendres formula which basically tells u the amount of prime factors a factorial has. 15! Is the answer cause when u divide 15 by 5 u get 3 factors of 5 and when u divide it by 2 u get more than 5 so u can see it’s the answer in less than 10 seconds.
Is it possible to do it faster by taking 5 as the prime factor? 1000 would become 5x200, 200 is 5x4, 4 is 2x2, so you will need two fives and two two's. This would also give answer D.
200 isn't 5x4, it's 5x4x10...and that extra 10 also breaks down to 2*5. So you end up with three 2's and three 5's, just like we discovered in the video. So the correct answer (D) must have at least three 2's and three 5's in it.
Your explanation is really excellent, currently doing my GMAT prep, and I want to ask, by tour method, if there's 10! at the answer list, would it be the correct chocie? Because we got three 5 from prime factorization of 5 and 10? Thanks in advance
I'm glad you found it helpful! No, as I explained around the ~10:20 mark, 10! would only have two 5's in it, so it won't be a multiple of 1,000. We need three 5's, so we have to bump up to 15! to get there. Good luck and let me know how else I can help!
Hey, quick question. I divided 1000 by 100 x 10 making less building blocks. You still get 5,5,5 and 2,2,2. If I use less building blocks, would this be a bad habit for other problems?
The question asks for the "least" value of N. So even if both 15! and 30! are divisible by 1,000, you need to choose 15 as the correct answer since it's smaller than 30.
It's still relevant! And rest assured that our course has been fully updated for the new GMAT Focus Edition if you decide to go that route. Good luck and let me know how else we can help!
That is so overcomplicated... I figured it out in 10 seconds. To be able to divide by 1000 we have to get three 0s at the end of the number. Only 5 or 0 when multiplied can give 0 as unit digit. Then no matter what you multiply by, 0 stays. 15 has two 5s and one 0. At least three zeros 000. Done.
@@Smooth_Operator I mean 0 can not be gotten multiplying any other numbers except 5 and 0. And as if there are not only odd numbers, multiplying the whole row together where 5 is also present - it would definitely give a 0 in result. 1x2x3x4x5(got 1st zero)x6x7x8x9x10(got 2nd zero)x11x12x13x14x15(3rd one). Interesting, now I see that it is even more universal: it works not only to find minimal number of 0's, but EXACT number. No = Round(F/5)
Guys Let me know if this is a correct approach .. Let's check for number of trailing zero in all of the options 1!=1 4! => No 5s in factorial 9 => one 5s in factorial so divisible by 10 only 15 => three 5s in factorial so there will be 3 trailing zeros .. this is the smallest number divisible by 1000 And hence 15 is the answer
So... does this count as an easier GMAT question? I'm in high school and I got it right so I'm just confused about the difficulty. The time it took was 30-90 seconds I guess idk. Sorry, kind of sounds like I'm flexing (although it does feel nice to get a question correct :) )
It's actually considered fairly difficult. But you're right, in many cases students are better prepared for the GMAT the day they graduate from high school -- at least on the quant side -- than after they've graduated from college and been in the workforce for a while, since this type of math is still fresh for them. The GMAT has a shelf-life of 5 years, so you might consider knocking it out now and then using it to apply to business school when/if you're ready.
Prime factorization and divisibility questions are generally considered "hard" by the GMAT, which would put this in the 600+ / low 700 range most likely. How did you find it?
That's great! Looks like you're well on your way to getting these types of questions right on test day. Just be sure that you study ALL types of quant questions you may see, as many of them are more "reasoning"-based questions where the traditional approach may not serve you.
Nor should you! You have ~2:00 per question on the GMAT Quant section. Once you've honed your pattern recognition and learned to apply the different strategies for each type of problem, solving questions like this in under two minutes should be very doable. Good luck and let us know how else we can help!
U won't hv so much time in exam to experiment. . . . . . Choices A B & C are not possible very clearly. So we are left with choice D & E. In 15!, 5*10*15 gives 750. Now lets make it a smaller fight & then go for thr bigger one. Imagine it is 75 and we need to see divisibility by 100. Now 75*4= 300 or 75*8=600 or 75*12=900. Any of these are divisible by 100. The bigger fight: 3000 or 6000 or 9000 are divisible by 1000. Hence answer is choice C. 😊
There is three terminating zero that means there should be three pair of 2 and 5 and and for count numbers of 5 = 15/5 = 3 and no of 2 = 15/2 + 15/4 +15/8 =7 +3 +1 =11 So ans must be 15! That's all. there is simple tricks
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I am quite close to my aim so I don't feel that I can justify buying your full pack, but your free stuff is amazing and whether or not I end up investing in the full quant I just want to say thank you so much for everything you have made available. It has been very useful and I'd easily recommend you to anyone I know.
Exactly, we really appreciate your free videos! god bless you!!
I got the ans as 15 and let me explain before I see your video for answer.
-I eliminated top 2 as none of the factorials are greater than 1000
-in 9 factorial, I searched for 3 unique combos whose multiplication results in multiples of 10. Only 5x4...hence ruled out
- In 15 factorial, I searched 3 unique combinations whose multiplications result is a muliple of 10 i.e.
1×10, 5x6, 15x2 ....
This means that the whole result set of 15! is divisible by (10x10x10) i.e. 1000
Hence, 15 is the answer
I do not simply listen to your lectures but also enjoy them! Thanks a bunch for your spectacular video. It adds to me a ton of motivation.
Thank you for the kind words. I'm glad it motivated you... and hopefully helped you better understand the principles to get similar questions right on test day!
Though I know this concept and got the question right, I was mesmerised by the patience with which you explained!
Thank you for the edification. Glad to hear that you're already up to speed on this type of question!
thank you! appreciate it! I'm currently facing a difficult period as my new job start date has been postponed few time, and I found your videos very helpful. This is the second one I'm watching.
Awesome, Chloel, I'm glad to hear it and thanks for sharing. I'm confident you'll find the other videos on this channel helpful and then whenever you're ready to dive even deeper, I hope you'll consider one of our paid courses. Best of luck and let me know how else I can help!
You are quite elaborate, thank you very much
My pleasure!
That was explicit. From this way of teaching, I deciphered easily.
Awesome, I'm glad it helped. I know you'll resonate with the way I teach other GMAT topics, too. I hope to see you in one of my courses!
Hey Brett. I was able to eliminiate answer choices A - C from the off. Just a question. Could we break down choice D / E as follows. 15! = (3 x 5)! Or 30! as (2 x 3 x 5)! and then solve.
wondering the same
cool, but i think it can't be solved like that cause (3 x 5)! is 720 while 15! is 1.307.674.368.000 .CMIIW
Thank you! Really helpful!
Very helpful! Thank you!
Thanks for great,detailed explanations..
My pleasure, glad it helped!
Use Legendres formula which basically tells u the amount of prime factors a factorial has. 15! Is the answer cause when u divide 15 by 5 u get 3 factors of 5 and when u divide it by 2 u get more than 5 so u can see it’s the answer in less than 10 seconds.
Nicely explained.
Thanks! Glad it helped.
This is a great method for sure
Is it possible to do it faster by taking 5 as the prime factor?
1000 would become 5x200, 200 is 5x4, 4 is 2x2, so you will need two fives and two two's. This would also give answer D.
200 isn't 5x4, it's 5x4x10...and that extra 10 also breaks down to 2*5. So you end up with three 2's and three 5's, just like we discovered in the video. So the correct answer (D) must have at least three 2's and three 5's in it.
Great explanation;).Thanks.
My pleasure. Glad it helped!
Learnt something. Thank You
You're welcome!
Your explanation is really excellent, currently doing my GMAT prep, and I want to ask, by tour method, if there's 10! at the answer list, would it be the correct chocie? Because we got three 5 from prime factorization of 5 and 10? Thanks in advance
I'm glad you found it helpful! No, as I explained around the ~10:20 mark, 10! would only have two 5's in it, so it won't be a multiple of 1,000. We need three 5's, so we have to bump up to 15! to get there. Good luck and let me know how else I can help!
That's quite interesting but won't it cost us a lot of time? I mean, time per question is approximately 2 minutes.
+Argirios Venekes Not now that you know how to do it!
Yes, it did! In fact, that was too much time spent on one problem.
Thank you! That was very helpful!
Thank you sir
You're welcome!
Hey, quick question. I divided 1000 by 100 x 10 making less building blocks. You still get 5,5,5 and 2,2,2. If I use less building blocks, would this be a bad habit for other problems?
Yes, it could get you into trouble on other questions because you need to know the minimum number of each prime factor you need in the numerator.
Thanks
My pleasure!
Why can’t we choose the option E as per my understanding it will also have the same factorisation like 15 could you explain me that?
The question asks for the "least" value of N. So even if both 15! and 30! are divisible by 1,000, you need to choose 15 as the correct answer since it's smaller than 30.
Great stuff!
Thanks David!
15/5 = 3 , 15/5^2 will not be taken as 5^2 is greater than 15, Hence min 5's are 3. 2's will always be more than 5's, hence direct answer is 15!
Thank you, thank you!!
You're welcome!
Watching this in 2024, really helpful
It's still relevant! And rest assured that our course has been fully updated for the new GMAT Focus Edition if you decide to go that route. Good luck and let me know how else we can help!
That is so overcomplicated... I figured it out in 10 seconds. To be able to divide by 1000 we have to get three 0s at the end of the number. Only 5 or 0 when multiplied can give 0 as unit digit. Then no matter what you multiply by, 0 stays. 15 has two 5s and one 0. At least three zeros 000. Done.
Nice way to think about it. Thanks for sharing!
what you mean "no matter what you multiply"?, 5x5=25, there are no 0's, i don't get it
@@Smooth_Operator I mean 0 can not be gotten multiplying any other numbers except 5 and 0. And as if there are not only odd numbers, multiplying the whole row together where 5 is also present - it would definitely give a 0 in result. 1x2x3x4x5(got 1st zero)x6x7x8x9x10(got 2nd zero)x11x12x13x14x15(3rd one). Interesting, now I see that it is even more universal: it works not only to find minimal number of 0's, but EXACT number. No = Round(F/5)
Guys Let me know if this is a correct approach ..
Let's check for number of trailing zero in all of the options
1!=1
4! => No 5s in factorial
9 => one 5s in factorial so divisible by 10 only
15 => three 5s in factorial so there will be 3 trailing zeros .. this is the smallest number divisible by 1000
And hence 15 is the answer
That's a good way to think about it!
Thank you very much sir! It’s 5am here, just going to bed
You're welcome. Glad you found us! (Get some sleep, it'll help you concentrate better on the GMAT...!)
Awesome
So... does this count as an easier GMAT question? I'm in high school and I got it right so I'm just confused about the difficulty. The time it took was 30-90 seconds I guess idk. Sorry, kind of sounds like I'm flexing (although it does feel nice to get a question correct :) )
It's actually considered fairly difficult. But you're right, in many cases students are better prepared for the GMAT the day they graduate from high school -- at least on the quant side -- than after they've graduated from college and been in the workforce for a while, since this type of math is still fresh for them. The GMAT has a shelf-life of 5 years, so you might consider knocking it out now and then using it to apply to business school when/if you're ready.
wait a minute, but 9 ! is 9*8*7*6 (at this point is = 3,024)*5*4*3*2*1= 362,880 which can be divided by 1000, then the answer is C, isn't???
Not evenly.
9!/1000=362.880
the question didn't say divisible evenly nor that the out come had to be an integer. please explain
@@HoustonMortgageTips Divisible definition (Webster) : (of a number) capable of being divided by another number without a remainder.
What level of Q is this? 500+? 600+? 700+?
Prime factorization and divisibility questions are generally considered "hard" by the GMAT, which would put this in the 600+ / low 700 range most likely. How did you find it?
🥴 i have done it in less than a minute 😂
That's great! Looks like you're well on your way to getting these types of questions right on test day. Just be sure that you study ALL types of quant questions you may see, as many of them are more "reasoning"-based questions where the traditional approach may not serve you.
I’m not spending 12 min on a GMAT problem
Nor should you! You have ~2:00 per question on the GMAT Quant section. Once you've honed your pattern recognition and learned to apply the different strategies for each type of problem, solving questions like this in under two minutes should be very doable. Good luck and let us know how else we can help!
U won't hv so much time in exam to experiment.
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Choices A B & C are not possible very clearly.
So we are left with choice D & E.
In 15!, 5*10*15 gives 750.
Now lets make it a smaller fight & then go for thr bigger one.
Imagine it is 75 and we need to see divisibility by 100.
Now 75*4= 300 or 75*8=600 or 75*12=900. Any of these are divisible by 100.
The bigger fight: 3000 or 6000 or 9000 are divisible by 1000.
Hence answer is choice C. 😊
There is three terminating zero that means there should be three pair of 2 and 5 and and for count numbers of 5 = 15/5 = 3 and no of 2 = 15/2 + 15/4 +15/8 =7 +3 +1 =11
So ans must be 15! That's all. there is simple tricks