[maxximus]

Hi all...this thread is dedicated to post key concepts to tacle specific type of problems...we all have our forte...n our grey areas...lets post well known ways as well as self-discovered ways to tacle problems we're comfortable with...this way, we'll get to learn thru each other...so if u have a specific way of dealing with a particular problem...that makes a difficult problem look easy...please let everybody know ur way...

we can also send our old posts, posts of other puys, info from net, or any means...the idea is to share concepts...no matter where they come from or when were they posted....as concepts are never old...those unknown are always new to us!

newbies and trainees can find lotsa info here...and i'd sincerily request the >=experts to come up with concepts gained thru their vast experience...

i'll post a concept almost everyday...and will also post few related questions (with answers to match)....hope this wud help us work like a gud team and we'll end up being better together.

warm regards
maxximus

p.s. special thanks to skylark who gave me this idea!


For those who are joining this thread late, here are the most important posts with important concepts...

post# 1,18,31,32,35,41,52,59,60,76,80,103,112,120,127,12 9,132,145,163, 192,193,226,246,250,314,322,323,325,344,345,346,34 9,355,372,390,404,410,411,412,414.

[pryank]

Hey yes u r right.. In this way round, of team work and sharing, we can grow to our mind and concepts.. and finally tackle CAT face to face... I'll be there wid u.... I hav an immense interest in Quant... Let us proceed...

[maxximus]

Today's funda...it's a combination of few of my old posts...


factorial based questions asking no. of zeroes and max power of sum integer.

Find the no. of zeroes at the right end of 300!

for every zero, we require 10..n every 10 is made up of 5x2.
in the expression 1x2x3...300, multiples of 2 wud obviously be more than the multiples of 5...so v need to find the maximum power of 5 in 300!

300/5 = 60 (because every fifth no. is a multiple of 5)

300/25 = 12(because every mutiple of 25 has two 5s in it) or, 60/5=12

300/125 = 3 (because multiples of 125 have three 5s in it) or,
12/5 = 2

now 2 cannot be further divided by 5 so add all the quotients...60 + 12 + 2 = 74.

we might also get the same type of questions in a different form,

500! is divisible by 1000^n...what is the max. integral value of n?

now every 1000 is made up of 3 5s and 3 2s....2s are redundant...we need to count no. of 5s....so find total no. of 5s and divide by 3

500/5 = 100
100/5 = 20
20/5 = 4

100 + 20 + 4 =124

124/3 = 41.33

max integral value is 41.


500! is divisible by 99^n...what is the max. integral value of n?

now every 99 is made of two 3s and one 11. obviously 11 will be the deciding factor. so count no. of 11s for the answer

500/11 = 45
45/11 = 4

ans will be 49.

so in such questions, just check which prime no. will be the deciding factor and count the no. of times it occurs. but please understand that highest prime no. is not necessarily always the deciding factor. see this example:

100! is divisible by 160^n...what is the max. integral value of n?

now 160 = 2^5 * 5^1. now although 5 is the biggest prime no. that 160 is made of, the deciding factor wud be 2. because five 2s occur less often than one 5 does. so we'll count the no. of 2s and divide by 5.

100/2 = 50
50/2 = 25
25/2 =12
12/2 = 6
6 /2 = 3
3/2 = 1

add 'em all...97.

97/5 = 19.

so the answer wud be 19

had v taken 5 as the deciding factor, the answer wud have been 100/5 + 100/25 = 24 which is more than 19...hence a wrong answer...

when in dilemma as to which prime no. wud be the deciding factor (e.g. a divisor like 144...its not possible to decide whether 3 or 2 will give the right answer) ....take out answer using both the prime nos...the one thats less is the right answer.

50! is divisible by 144^n...what is the max. integral value of n?

144 = 2^4 * 3^2...difficult to decide whether 3 or 2 will be the deciding factor...

count 2s

50/2=25
25/2=12
12/2=6
6/2=3
3/2=1

sum=47

answer = 47/4 = 11.

count 3s

50/3=17
17/3=5
5/3=1

sum = 23

23/2 = 11

a tie...else the smaller value wud have been the answer.

300! is divisible by (24!)^n. what is the max. possible integral value of n?


such questions are tricky...when u expand 24!...u get 1x2x3...24.

in this range the highest prime no. is 23...so maximum power of 23 in 300! will decide the max value of x...

when v expand 300!...v get a 23 in 23, 46,69,92....

total no of multiples of 23 in 300! will be 300/23 = 13,

forget the fractional part. so the maximum possible answer is 13. hope am clear...else, feel free to revert.


256! is expanded and expressed in base 576 . how many zeroes will this expression have on its right end?


such questions are same as finding maximum power of 576 in 256!

576 = 2^6 x 3^2
to get six 2s i have to travel eight places...1x2x3x4x5x6x7x8 has seven 2s. but to two 3s i have to travel only six places...1x2x3...6 has two 3s...hence 2 will be the constrain.

total 2s in 256! = 255

hence, no. of zeroes = 256/6 = 42.

just to check...3s = 126, 126/2 = 63>42

ans-42




Questions based on this concept

400! is divisible by x^n. what is the max. possible integral value of n if the value of x is:

Q1. 300
Q2. 99
Q3. 500
Q4. 320
Q5. 770
Q6. 5200
Q7. 270
Q8. 686
Q9. 338
Q10. 13000

Answers... 49, 39, 33, 66, 39, 32, 65, 22, 16, 32 (the answer is not 33, this one is actually tricky! )

200! is divisible by (x!)^n...whats the max. possible value of n when x =

Q11. 25
12. 35
13. 50
14. 100
15. 70
16. 300
17.15

answers... 8, 6, 4, 2, 3, 0, 16

300! is expanded and expressed in base x. find the no. of zeroes at the right end of this expression when x=

18. 25
19. 15
20. 35
21. 39
22. 98

answers...37, 74, 48, 23, 24

do lemme know if there's any problem at all.

cheers!

maxximus

[iyervani]

Originally Posted by maxximus

Today's funda...it's a combination of few of my old posts...


factorial based questions asking no. of zeroes and max power of sum integer.

....
maxximus

hey maxximus....very very useful .....Thanks a ton!!!! do keep them coming....
this thread is going to go a long way!!!!! i am sure!!!

[GuruBen]

Originally Posted by maxximus

Today's funda...it's a combination of few of my old posts...


factorial based questions asking no. of zeroes and max power of sum integer.

Find the no. of zeroes at the right end of 300!

for every zero, we require 10..n every 10 is made up of 5x2.
in the expression 1x2x3...300, multiples of 2 wud obviously be more than the multiples of 5...so v need to find the maximum power of 5 in 300!

300/5 = 60 (because every fifth no. is a multiple of 5)

300/25 = 12(because every mutiple of 25 has two 5s in it) or, 60/5=12

300/125 = 3 (because multiples of 125 have three 5s in it) or,
12/5 = 2

now 2 cannot be further divided by 5 so add all the quotients...60 + 12 + 2 = 74.

we might also get the same type of questions in a different form,

500! is divisible by 1000^n...what is the max. integral value of n?

now every 1000 is made up of 3 5s and 3 2s....2s are redundant...we need to count no. of 5s....so find total no. of 5s and divide by 3

500/5 = 100
100/5 = 20
20/5 = 4

100 + 20 + 4 =124

124/3 = 41.33

max integral value is 41.


500! is divisible by 99^n...what is the max. integral value of n?

now every 99 is made of two 3s and one 11. obviously 11 will be the deciding factor. so count no. of 11s for the answer

500/11 = 45
45/11 = 4

ans will be 49.

so in such questions, just check which prime no. will be the deciding factor and count the no. of times it occurs. but please understand that highest prime no. is not necessarily always the deciding factor. see this example:

100! is divisible by 160^n...what is the max. integral value of n?

now 160 = 2^5 * 5^2. now although 5 is the biggest prime no. that 160 is made of, the deciding factor wud be 2. because five 2s occur less often than one 5 does. so we'll count the no. of 2s and divide by 5.

100/2 = 50
50/2 = 25
25/2 =12
12/2 = 6
6 /2 = 3
3/2 = 1

add 'em all...97.

97/5 = 19.

so the answer wud be 19

had v taken 5 as the deciding factor, the answer wud have been 100/5 + 100/25 = 24 which is more than 19...hence a wrong answer...

when in dilemma as to which prime no. wud be the deciding factor (e.g. a divisor like 144...its not possible to decide whether 3 or 2 will give the right answer) ....take out answer using both the prime nos...the one thats less is the right answer.

50! is divisible by 144^n...what is the max. integral value of n?

144 = 2^4 * 3^2...difficult to decide whether 3 or 2 will be the deciding factor...

count 2s

50/2=25
25/2=12
12/2=6
6/2=3
3/2=1

sum=47

answer = 47/4 = 11.

count 3s

50/3=17
17/3=5
5/3=1

sum = 23

23/2 = 11

a tie...else the smaller value wud have been the answer.

300! is divisible by (24!)^n. what is the max. possible integral value of n?


such questions are tricky...when u expand 24!...u get 1x2x3...24.

in this range the highest prime no. is 23...so maximum power of 23 in 300! will decide the max value of x...

when v expand 300!...v get a 23 in 23, 46,69,92....

total no of multiples of 23 in 300! will be 300/23 = 13,

forget the fractional part. so the maximum possible answer is 13. hope am clear...else, feel free to revert.


256! is expanded and expressed in base 576 . how many zeroes will this expression have on its right end?


such questions are same as finding maximum power of 576 in 256!

576 = 2^6 x 3^2
to get six 2s i have to travel eight places...1x2x3x4x5x6x7x8 has seven 2s. but to two 3s i have to travel only six places...1x2x3...6 has two 3s...hence 2 will be the constrain.

total 2s in 256! = 255

hence, no. of zeroes = 256/6 = 42.

just to check...3s = 126, 126/2 = 63>42

ans-42




Questions based on this concept

400! is divisible by x^n. what is the max. possible integral value of n if the value of x is:

Q1. 300
Q2. 99
Q3. 500
Q4. 320
Q5. 770
Q6. 5200
Q7. 270
Q8. 686
Q9. 338
Q10. 13000

Answers... 49, 39, 33, 66, 39, 32, 65, 22, 16, 32 (the answer is not 33, this one is actually tricky! )

200! is divisible by (x!)^n...whats the max. possible value of n when x =

Q11. 25
12. 35
13. 50
14. 100
15. 70
16. 300
17.15

answers... 8, 6, 4, 2, 3, 0, 16

300! is expanded and expressed in base x. find the no. of zeroes at the right end of this expression when x=

18. 25
19. 15
20. 35
21. 39
22. 98

answers...37, 74, 42, 23, 21

do lemme know if there's any problem at all.

cheers!

maxximus

This thread is going 2 be v useful.
However, I am not v clear on how do we chosse the deciding prime factor.

In the case if 160 = 2^5 * 5^2.......we chose 2 as the deciding factor. I am assuming it because a 2^5 > 5^2
or is it something else???????

In the eg of 300! is divisble by (24!)^n we chose 23 . However wabt abt 2. There are 2^10 in 24!. Then howcome 2 is not the deciding factor(going by my previous assumption)

Do clear my doubt.

[vallisri]

based on concept above:
i want to know wht will be the answer to the question:
how many 24's are in 150!

24=2^3 * 3
My answer is 48
as we get number of 2 in 150!=145
then 145/3=48

so is this answer correct or any mistake in procedure ...do lemee know.

[The0ne]

refer arun sharma this funda is present in it

[maxximus]

hi guys...thanx for the interest being shown here...keep it up...keep pouring in ur doubts n concepts...

@vallisri... no. of 2s - 48

no. of 3s - 45

but 24 = 2^3 x 3^1

that means, v need three 2s and one 3 to get a 24.

now see...1x2x3x4x5x6x7x8...24.

to get one 3, v need to travel only 3 places...viz 1x2x3.

but, to get three 2s, v need to travel 4 places...viz 1x2x3x4.

that means three 2s occur less frequently than one 3 ... so the constraint wud be 2.

to get the correct answer, divide 48 by 3 to get 16.

or simply, since 48/3 < 45/1, the answer shud be 48/3 = 16.

try few of the problems mentioned above...if u get 'em right...concepts are in place...
feel free to revert



@guru...please delete contents of original post except the relevant part to avoid the useless length of post and ease of understanding the doubt/concept.

originally posted

In the case if 160 = 2^5 * 5^2.......we chose 2 as the deciding factor. I am assuming it because a 2^5 > 5^2
or is it something else???????

In the eg of 300! is divisble by (24!)^n we chose 23 . However wabt abt 2. There are 2^10 in 24!. Then howcome 2 is not the deciding factor(going by my previous assumption)

Do clear my doubt.


1...slight typing error mam...160 = 2^5 x 5^1...i've mentioned it later that five twos occur less frequently than one one 5.

to get five twos...v need to travel six places...viz...1x2x3x4x5x6

to get one five....v need to trvael jus 5 places...1x2x3x4x5.

so constraint is 2. find total no. of 2s and divide by 5. in case its difficult to say which prime no. wud occur less often get answer thru both the prime nos...smaller of the values wud be correct answer.

the approach shud be...

(no. of times a prime no. occurs) / (no. of times its required to form divisor)

here in 160!... 2 occurs 80 + 40 + 20 + 10 + 5 + 2 + 1 = 158 times

158/5 (because 160 = 2^5 x 5^1) = 31

in 160!, 5 occurs 32 + 6 + 1 = 49 times.

since 31 < 49, answer is 31.


2. please understand that u dont have to match absolute values...u need to to check which no. wud be the constraint....am sure u have not read the post properly...else u wont have said there are are 2^10 in 24!....there are 2^22 in 24!...but so wat...there are 2^296 in 300!...taking 2 as constraint wud gimme 296/22 = 13.45....which is more than 300/23 = 13.04..although the answer wud be 13 in both case...the point am trying to make is in such questions, dont be prejudiced...check which prime no. wud be constraint...dont take 2^5 as 32...think that to get a 32, 1x2x3x4x5x6 7x8 is more than enough...but to get a 23...we need to go a longer way..1x2x3...23


read the posts 2-3 times....u'll understand...n then attempt the problems given by me...check ur answers...dont doubt them...they're all correct...keep trying u'll slowly master the concepts

atb!

@theone...thanks for the interest shown here bro...but please understand...this thread is not meant for bookish concepts...we dont refer books here...v discuss our practical...short cut methods...with due respect to the author...we know a.s is gud for practice...not for concepts...we'll appreciate if u cud share ur methods gathered from sources instead of blatantly refering the source...no offence...please take this in the right spirit...the thread is meant for practical approaches...take gud care n please keep posting


regards
maxximus

[vallisri]

Originally Posted by maxximus

hi guys...thanx for the interest being shown here...keep it up...keep pouring in ur doubts n concepts...

@vallisri... no. of 2s - 48

no. of 3s - 45

but 24 = 2^3 x 3^1

that means, v need three 2s and one 3 to get a 24.

now see...1x2x3x4x5x6x7x8...24.

to get one 3, v need to travel only 3 places...viz 1x2x3.

but, to get three 2s, v need to travel 4 places...viz 1x2x3x4.

that means three 2s occur less frequently than one 3 ... so the constraint wud be 2.

to get the correct answer, divide 48 by 3 to get 16.

or simply, since 48/3 < 45/1, the answer shud be 48/3 = 16.

try few of the problems mentioned above...if u get 'em right...concepts are in place...
feel free to revert

regards
maxximus

Thank you maxximus,
But i did not understand why did you divide by 3.
Could u plsss explain.

[maxximus]

Originally Posted by vallisri

Thank you maxximus,
But i did not understand why did you divide by 3.
Could u plsss explain.

24 = 2^3 x 3^1

so v requir 2s to be in sets of three. if v have 48 2s...how many groups of three twos are possible? 48/3 = 16.

please read the main post few times...this is an important concept n shud be mastered....be in no hurry...dont jump to conclusions....there are many cases...i've discussed them all...read carefully.

regards