Factorials 3 || Number Systems || Quantitative Aptitude ||CAT Preparation 2024

Number System by Ravi Prakash | Quantitative Aptitude for CAT 2024
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[@AmitThakur-eg8kb]

Here is the detail explanation of q7. let x^2 = 5000 x = sqt(5000) x = 70.71 now here we need to consider x = 70, now the concept here is we need to check for the prime number greater then 70 . And it is 71 so check for 71 5000/71 = 70.42 71^70 * k/ 71^71 here you can see it is not completely divisible. so we need take all the prime numbers below 71 all the prime numbers below 71 is 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67. total 19 prime numbers .. 19 is the answer Let me know if you have any confusion.

[@Adityatulsyan]

Do we have to manually count how many prime numbers are before 71 ?

[@jainamhemanee1125]

Thank you bro.

[@shrutigautam9383]

@@Adityatulsyan or it's always better to know the prime numbers till 100

[@amishatiwari8941]

Please help me ..how to solve this 101^99*k/101^101 How we know it is dividing or not

[@bhavadharrani4829]

@@amishatiwari8941 101×101×............. 99th times divides by the same 101×101×..........101th times finally you will get 1 / 101×101 so it is not fully cancelled right..

[@pritheshnarang2166]

Sir, I have recently started seeing your videos and I believe these lectures are really a piece of gem for the people who are preparing for CAT and other exams. I'm glad that I found your channel.

[@Yuvaraj5-lc6yy]

How can we written 50!= 49!X 50 👈🏻👈🏻🙏🏻

[@clutchlabs]

5!= (1*2*3*4)*5=4!×5 similarly

[@samirbehera-k7k]

the answer of question 7 will be 19

[@AshutoshKumar-px8wo]

Sir your factorials vedio 2 isnt running please solve that

[@meghamehta9492]

Please someone tell me why 4 is also included in 5th question.

[@thechannel3420]

It might be just divisible, not completely divisible. 2 × 2 = 4, one 2 is divisible

[@cockopuff]

@Akash bro thank you so much i mean this is the doubt one should ask but people are askin such silly questions .

[@sanyamsinghi1802]

u are donating crores !❤

[@ankit1666]

Who else is preparing for 🐱🐱🐱🐱🐱 2024

[@kabirmanek7884]

Your method of teaching is amazing.....thankyou sooo much!...…..ans to 7th question is 19.

[@gauravrajchouhan4756]

answer should be 15.. kyuki 15 prime number hote hai less than 49

[@vishvesh10]

@@gauravrajchouhan4756 Nope bro, you're not following the concept. => x^2=5000 => x=70 not 50. We'll consider all the prime numbers before 70, that is 19.

[@027_manishdixit7]

19

[@senthilnkp1281]

Got the same

[@karanjain2359]

@@senthilnkp1281 correct

[@arghadeepdas520]

This was a much-needed channel, at this time!

[@ayushsrivastava9947]

Q7. Find nearest sq. of 5000 which is smaller than or equal to 5000, here we can take 70, as 70*70=4900. So it will be divisible by all prime numbers till 70, Hence 19 prime numbers. 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,51,59,63,67

[@dhruv20]

i dont get how in 10k!/p^p ,p is prime, question ,sir got the idea of starting with dividing 10k by x t and write x^2=10k

[@hiteshkhatwani2486]

51 and 63 are not prime no.s

[@AnanyaDoneria-te1ql]

@@hiteshkhatwani2486 that must be 53 & 61

[@narenpatil14]

25:02 sir shouldn't highest power of 97 be 104 10000/97 = 103 , 103/97 = 1.xyz... 103+1 =104

[@Diabolical3010]

Yup

[@studyyyy.commmm]

It's says only prime no. 104 is not prime

[@devendrasontakke1694]

Plaese improve the voice quality and sound

[@Jerryon60fps]

35:17 sir in this question 8, even if someone was able to catch the logic of trailing zeroes and they come to a conclusion that of course whatever the addition of the factorial is, its going to end with many 0's. shouldn't Tens place be 0 if we count? how will we be able to catch that we have to actually consider 1-9 digits in the number to count as the "tens digit" and ignore the many zeroes that it may have??

[@anshbhatia2740]

Thank you very much sir! Beautiful videos! And yes, we are pausing and trying xD

[@venkateshautogarage8437]

Dude, have you got answer for all the questions?

[@yudhistergaming4394]

Why have sir written 50! As 49!×50

[@yes3635]

@@yudhistergaming4394 50! Is what?? 1x2x3.......48x49x50= 50! Right? So we can write 50! As 49!x50=50!

[@NaveenRajha]

I have an idea of preparing for cat...it just suddenly stricked me just three days before...iam really poor in maths..hoping to give time to prepare for cat 2026 and 2027...even if i couldn't crack also.. atleast iam getting better in my mathematical skills...that is my end goal...

[@Nandana14312]

How's preparation?

[@coffeewaarushi]

I'll be giving cat 2024 to get the grasp of how the paper is like so that for cat 2025, I'm all prepared and set. I'll know that this isn't my first time i shouldn't be scared and that i can do it. That's what my wish is, let's see how it goes 🤞

[@iamdavidaman]

At 28:35 the solution for that question is 19 Explanation X² = 5000 Then x = 71 (not exactly) So check all prime no. That lies below 71 i.e-2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67 For P = 67 5000!/P^p = 67⁷⁵ x k / 67⁶⁷ For P = 53 5000!/P^p = 53⁹⁵ x k / 53⁵³

[@sumedhj7917]

But dividing 3! by 4 it leaves the remainder 2. So, is not it necessary to completely divide that number?

[@prabhavjain8391]

exactly

[@harshitabajaj2020]

here the exception must be 5! as contains 3x2 divide by 6

[@hussainpunawala6113]

12:00 in this qsn i have take out power of 5 in 250! Still i got 5^62

[@govindrituraj1816]

Just 2 months left and now I decided to give CAT and so I don't want to waste 40k for 2 month and here comes your channel just a blessing for me hope all u r teaching is sufficient for exam

[@mohdsheeraz4463]

was it sufficient?

[@foodie_byhaq9685]

Did u clear CAT?

[@divyajain6183]

how was your exam? is it ok if i self study by watching these vedios ?

[@ijatinghuman]

@@divyajain6183 these videos are better than my time coaching institute lecture.

[@X-Review]

Anyone preparing for CAT 2023

[@gamingclaw25]

Yes

[@GauravDutta-yk]

how in Q.3 (7^2)9= 98^9, Why not 49^9???

[@Khush_ghuman]

Who else is prep here for cat 2020

[@vikkii3202]

ME here

[@mohanadeepika8785]

Me

[@prateekparashar6917]

Me here

[@kunalmjain]

Me

[@XLBOX1]

Yahhhh there

[@MaxwellNaradd]

Ravi Sir at his earliest in these factorials videos. Absolutely Flawless!!!

[@ayanavamukhopadhyay5863]

Find the highest power of 5 in the product of first 50 multiples of 5. Another approach: Find the number of 5's in the factorial of 5*50(250), which is 62.

[@ankitwatts1918]

I also tried this, and is suited for any amnother questions too

[@arunmishra3691]

I am from IIT and trying to go to IIM. Trust me, this guy is a fantabulous teacher. Respect!!

[@sonacollegeofartsandscienc9247]

Which IIT ?

[@shramande9302]

IIT Pochinki ?? 🤔🤔

[@VarunKumar-jv4xt]

@@shramande9302 🤣🤣🤣

[@KrishnaSharma-op6io]

Are you getting in iim?

[@maverick9319]

🤣🤣🤣

[@prakashdesai4126]

Where can I get practice questions? If anyone has idea....please help. Thanks in advance!

[@mehakahuja5708]

From Arun Sharma book

[@Randomworldindia]

@@mehakahuja5708 try rs aggarwal it's harder then Arun Sharma and better for cat exam

[@ratanprajapati93305]

Chori chipe pada rhe ho kya itna volume kam hai tumhara

[@samirbehera-k7k]

the answer of question 7 will be 19

[@rashimohan314]

In q5, aren't all multiples of 4 exceptions? 11! is divisible by 12, 7! is divisible by 8 and so on...

[@mayurijadhav3372]

B'coz multiple of 4 cannot be prime

[@V_Kavya]

Sir in Q3 i.e., find highest power of 98, 7 goes (14+2) 16 times so final power is 8 na, how 9?

[@suhas8530]

Hi at min 18:00, I have some confusion. The question asks for no. of natural no's less than 200 not divisible by n. Is the solution correct? can anybody explain?

[@krishnasandhya7230]

Answer for 7 th question is 19 (p

[@AakroshKalsotra]

no its all no less than 50 ie 15

[@etukurudinesh1926]

when p=97 no of 97's in 10,000! is 104. But you are using 103 in ur lecture. hence it is 97^104. in this case it doesn't really matter b/c we are just checking to see the (power in numerator > power in denominator). but i just want to point it out.

[@anilbhai1652]

Good observations, i too noticed that.

[@mansikhatak1258]

In Q8, if we add all the values i.e. 1+2+6+24=33 , then at ten's place, shouldn't it be 3 , kindly correct me if I am wrong

[@shibbufc4163]

Answer of question no. 7 is 19 Thanks for the method sir ..

[@vilegend9339]

How bro

[@theuntoldrecipes4322]

Your 2nd class of factorial is not running plz do something

[@harshthakkar6653]

Yes

[@adeshpatil4792]

play on 360p

[@YineeTV]

I am also human I can also do mistakes 😜😜

[@GauravUP16]

sir video m voice kam hai

[@akshatsinghOG]

did not understand Q5

[@ankeshsingh2576]

We could have calculated the tens digit with the sum of 1! to 4! as well, as it generates a carry and from 5! onwards every other factorial with have one zero atleast. Isn't it ?

[@kshitijbalonie4361]

in ques no. 6 shouldnt it be 97^104 rather than 97^103?

[@deepshikhachaubey4189]

Sir as the question asked 1!*2!*3!*4!........*30! Are we not suppose to multiply the trailing zeroes of each factorial as in starting from 10! 2*2*2*2*2*3*3*3*3*3*4*4*4*4*4........*7 here in this solution we added the trailing zeroes. Can you please clear my doubt.

[@vivekpandey7824]

Sir ! The line n=4. 3! 4 exception..I don't get this..pls explain..

[@pranjalipethkar4026]

n=4, then (n-1)!/n= 3!/4 = 6/4, not divisible. that's why 4 is the only composite number in this question that is an exception

[@kamalkishorebajaj143]

@@pranjalipethkar4026 thanks

[@yoga1freak]

Sir, in that prime numbers below 200 question, even if we take the exception of 4, still total numbers should be 46 as we will have to remove the case of (1-0)!=0!=1, which is divisible by one. Therefore, 46(prime numbers +1(Case of 4) - 1(case of 1).

[@Singhaszn]

1 is not a prime number

[@anishchar4282]

Who else is prep here for cat 2023

[@Movewithkhu]

True for all prime numbers 1-200 exception of 4 the factorial won't be divisible by n-1

[@iamdavidaman]

at time 10:15 can we solve this question like 5x50 = 250 and if we calculate the number of 5 in 250! = 62 m i right or wrong ?

[@bhavyamanchanda7840]

sir didn't understand question no 8...i didn't understand the concept behind it .....would request u to explain again ...

[@rishisrivastava5461]

sir please do upload practice sets..

[@apollodavis4090]

q6 went over my head

[@ronakdubey4088]

🤣🤣

[@abhishekdebnath7112]

you provide the best class out here in youtube. thank you very much for clearing my concepts.

[@raviprakash6021]

do join our WhatsApp group for doubt clarification and LRDI Concepts & Practice.. chat.whatsapp.com/HQQvw5IWFHz4IwLbS7pojS And FB group for question bank .. facebook.com/groups/570194513099989/

[@ipsitahalder2981]

@@raviprakash6021 Sir, the whatsapp group is full. Are there any other whatsapp groups where you provide guidance?

[@sparshkalra9439]

Sir dubu still have a watsaap group ?

[@sakshamlanjewar5523]

How is 3! Divisible by 4?

[@alinjardas1861]

it is not divisible by 4, the question wants numbers not dividing n-1!

[@sharayupatil1424]

Sir ...you are doing a great great job ....thank you soo much for solving so many questions ....making us understand the concept right from its core and putting soo much efforts🙏🙏

[@ashishchaturvedi222]

sir at 18:34 if we take n = 4 then, we can't completely divide 3! by 4 so the answer has to be 46, as there are 46 prime numbers till 200 natural numbers. pls clear my doubt

[@chinmoyprasad5914]

You are correct Ashish. 3! = 6, so not divisible by 4. If there is an exception it must be something other than 4, and idk how to check that. So I guess the answer should be 46.

[@yashajain6200]

I have the same doubt how come 4 is an exception

[@harshvardhan8261]

yeah the ans is 46 .

[@sidhantpatra1652]

Question is not asking for No of prime numbers. Question is asking for no of natural numbers so 4 should be included and the answer will be 47 (46 prime numbers+ 4 )

[@josephbj8766]

For Q6, (10000!/ 97^97) = 5.46 Hence it is not completely divisble right?

[@nahishgupta4446]

Yes

[@shubhankarsarkar1995]

Thank you Sir for these challenging problems and their never heard before concepts!!

[@techincalsketch5686]

Sir i have a doubt...how to solve this 25!to 29! = 6 sir plz tell me it's a humble request?

[@spkbusinessD--]

shivam kamat(6) 1-8-2021 thank you sir 🙏

[@parkhipriyadarshi8342]

3:58 .. Hi , Can you please explain it how it is 49! ×49!..

[@Ok12376]

HOW TRUE IS THAT. AWESOME. 9! IS 3,62,88'0' 10! IS 36,28,8'00' 11! IS 3,99,16,8'00' 12! IS 47, 90, 01, 6'00' 13! IS 622, 70, 20, 8'00' 14! IS 4717, 82, 91, 2'00' AND GOING ON.

[@deenaselcia1459]

Sir in 5th question 2 is there in 3! so it is divisible by 4 then y ur not considered that

[@rashibhatnagar6772]

In ques no.5 , is not 8! divisible by 9 so why have we taken only 4 as exception and not included 9 in the answer?

[@sonacollegeofartsandscienc9247]

9 is not a prime number. And 9 can written as 3×3. It has factors. 8! Can be divided by 3×3. 1×2×3×4×5×6×7×8/3×3 3 and 3 canceled and the remaining 3 in the denominator is cancel with the 6 in the numerator. So this is divisible. It doesnt fulfil the condition

[@rashibhatnagar6772]

@@sonacollegeofartsandscienc9247 thankyou

[@shwetalpatel6]

sir where is AP GP?

[@mindfullmallu7511]

Tq sir. U are unique 🥺❤️

[@saurabhchandra4320]

Sir, question number 6 here. 97*97 how this is completely divisible with 2499 trailing zeroes ? can you give one explanation ?

[@alexandersupertramp3415]

Completely divisible means there's no denominator left

[@mayankmalik9818]

hello sir! Didn't get question no. 7th. Would request you to tell in an easy way.

[@anilbhai1652]

First focus on how did you get x = 100, just by square rooting of 10000, and finally find the no. of prime numbers Which are less then 100, so likewise you have to do square root of 5000, and answer will be 70, now you have to find how many prime numbers there are less then 70, answer will be 19.

[@ThatMBAstudent]

12:07-18:30

[@Ashish-wf7tb]

Q3 7^2=49

[@vyommaheshwari7130]

bhai mere audio theek karde videos ka. pankha chala ke vidos suno lo toh aisa lagta hai behre hogye hai

[@Littlerose1011]

19 ?

[@sanjeevgautam6391]

sir you are amazing .you are doing a great job and helping students learn for free ..i will recommend you to all my friends who are preparing for cat. thanks once again.

[@krishnagohil2983]

Guys can anyone recommend me any ebook or guide for cat

[@mayankpandey9752]

factorial concepts are quite lengthy

[@akwork2483]

Q5. what if the question says completely divisible by n. in such cases will 4=n be an exception ?

[@amitpatel4605]

Same doubt

[@deepeshparihar7362]

it says how many number less than 200 cant where ( n-1 ) ! is ' not ' divisible by 'n' . So in 4 case ( 4-1) 3! will be 6 and its not divisble by 4. But in 6 5! is 120 and it is divisible by 6. I also checked same in 8 case 8-1 = 7! is 5040 and it is divided by 8. So according to this all question and only 4 can not divide ( n -1 ) ! and if it itself is 'n'.

[@deepeshparihar7362]

All prime numbers can't divide ( n -1 )! and also 4. in this case 4 will be the exception because 4 is the only ' not prime ' number which cannot divide ( n-1) ! . Other non - prime numbers can divide ( n-1)! except for 4.

[@deependrapratapsingh2401]

Ques 7 answer is 19 28:42

[@meenakshitiwari736]

Sir question 3rd ka ans 8hona chaiye ..... Kuki ap jo kr re h uske hisab se 7ka square 49hota h to no of 49 niklra h pr hume 98chaiye .. To hume 2ko b consider krna hoga

[@shamgurav21]

as he told in this video no. of 2's are greater than no. of 7's. hence we don't need to find the occurrence of no. of 2's.

[@gosolo01]

have done ques no 6 5000!/p^p ? is the answer is 18

[@tusharpandey1872]

@@gosolo01 19

[@sukhmanikalra47]

in question 4, two 5's are needed to make 50 factorial.. so in ans why we didn't divide like we did in previous questions??

[@310manyadiwan8]

After seeing your comment, i am confused too.

[@nidz0109]

Same question

[@user-nr6uh3ld3m]

I have the same doubt

[@chandrakalajadhav8519]

In Q1) I can't get it because of sign of multiply within the factorial....what happen when there was plus sign.... ?

[@ayushsrivastava2397]

can any one plzz explain me the 6th ques...ppzz

[@rohitchakrabarty5684]

Who is preparing for CAT 2023

[@divyajain6183]

why we write 50!=49! x 50 i can't understand.....

[@eyaShukla]

It's 49! x 50 , not 50!

[@anureetjuneja2903]

For Q4, I simply calculated the factorial for 250 in terms of 5, and the answer came out to be the same. Is this the correct method?

[@sjsthomasnick3002]

How can we do it? Can you please explain the logic behind it?

[@sjsthomasnick3002]

@ENGINEER'S HOUSE can you please explain the logic behind doing this way?

[@paanda_p]

Yes I am also thinking about this and i solved this in the same manner

[@ayushdeshmukh7697]

@@sjsthomasnick3002 first write it as 250!-4! But as there is no multiple of 5 in 1! to 4!, We can directly calculate the number of 5 in 250!. This is a simpler method Also I don't know if I'm making sense but I've understood it

[@Shubham-wv6tf]

​@@sjsthomasnick3002😅

[@zalakjani6212]

sir can you please explain why did we wrote 7 raised to 10 as 49 raised to 5

[@Deardiaryyyyyy]

hey because 49 is the result of multiplying 7 twice ,hence we are taking 2 sevens for one 49, meaning 49 raised to 5 and not 10

[@amirthavarshni5718]

Hi Sir! Ur channel is very helpful for my preparation. I just have a doubt in the 5th question. U said that 4 is the only exception. But isn't all the squares of the prime numbers an exception? Because, in the case of 49 also 48 factorial is not divisible by 49 naah? And 49 is neither a prime number. So we should include the squares of the prime numbers too into this account no, sir?

[@Mindful_banker]

No , 48! Is divisible by 49.

[@faraznadeem7918]

Well! In this case it is asked that n should not divide factorial (n-1)!. Considering n=49 (N-1)! = 48! ( Remember its 48 factorial not only 48) I.e 1! *2! *3! *4! *5! ............48! Spliting 49 into prime numbers it will be 7*7 which will eventually divide 7,14 ,21,28,35,42 therefore 48! is divisible by 49. Hope you got it . 😉

[@nikitapardhi3438]

@@faraznadeem7918 hey can you please explain question number 3 in detail...

[@Randomworldindia]

@@nikitapardhi3438 See 98 = 49×2 We can also write it as = 7^2 × 2 Here just ignore 2 ok So 98= 7^2 now Now we have to find how many 7's are there in 98! So now We can take 98 as common from both side Therefore, =98! (99-1) = 7^2 × 98! = 7^2 × 7^16 =7^18 we can write again 7^2×9 =(49)^9

[@nikitapardhi3438]

@@Randomworldindia hey thanku for your explanation...

[@akritibaranwal9081]

Sir in question no. 1 why 7 power 10 is taken as 49 whole power 5?

[@dhruv20]

i dont get how in 10k!/p^p ,p is prime, question ,sir got the idea of starting with dividing 10k by x t and write x^2=10k

[@ganeshangonda2132]

For q7 can anybody take value of 1000 instead of 5000 and solve please?

[@sjsthomasnick3002]

Square root of 1000 is approximately 31. So all prime numbers from 1 to 31 will safisfy the condition 1000!/p^p . 2,3,5,7,11,13,17,19,23,29,31 There are 11 prime numbers from 2 to 31. Ans : 11

[@saikartheek992]

Sir 7^2 is 49 ..then how can you say thay the answer is 9 for question no.3

[@abeginnerspov5968]

99!-98! = 99*98!-98! = 98!*98 = 7^16 * 7^2 * 2 = 7^18 * 2 = 49^9 * 2 = 98^9. We considered 7 because 7 is the highest 1 digit prime number, and higher is the base number, lower is the power. Other than that, 7*7*2=98. We want the number of 98's.

[@rishipandey_006]

how to breakdown 98 anyone know..

[@country.mirror]

Do it mentally yrrr 😂😂

[@arijeetpandafit]

about Q no 4 - i have done by finding out multiple of 5 in 250 ! . how i get 250 ! ? sol - as given multiple of 1st 50 so in 10 numbers their are 2 multiples of 5 . thus , 10 * 50 / 2 = 250! is it can be a correct way ?

[@sakshisharma4694]

19 is the answer of ques 7

[@gumnaam4117]

Ye episode kuch samaj nahi aaya

[@suryavoroy3527]

Tricky question at 23:00

[@sidharthsharma3158]

sir q5, why all the multiples of 4 can't be divisible by n-1 factorial? like 8,16

[@Prithvi-s3j]

One question on 10000!/p. When you say a number is completely divisible then rem should b 0 right.but all the prime num b4 100 wont giv rem 0.

[@kshitijbalonie4361]

Thankyou sir

[@gdavid7090]

@Rodha sir..... in q5......... if n = 10...... (n-10)! = 9! is divisible 10 ..... it is also an exception............. please explain sir

[@dhruv20]

i dont get how in 10k!/p^p ,p is prime, question ,sir got the idea of starting with dividing 10k by x t and write x^2=10k

[@shailendridewangan7353]

Thankyou so much . This is the best KZbin channel for all cat aspirants .