[maxximus]

Originally Posted by junoonmba

Hi maxximus!!!!!!!

Few corrections required in your post
1)first answer of 10 should be 16 not 30(In bold ,because the logic and explanation given by you itself says so......)

2)15! has 6 3's not 7 thus multiply factor should be7 (6+1) not 8.....
15/3=5
5/3=1

I stll think answer of 17) should be 60..(30+30),ponder a lot but culdnot found themistake.....

sorry yaar..i thought the question was 17.positive and 18. negative...ur answers are correct...gud job...sorry for inconvenience.

regards
maxximus

[maxximus]

Today's concept....finding squares and close multiplications quickly...

In a hurry guys....so few small concepts that'll help u save sum time...n avoid cramming....which i've always maintained is the best ill-preparation for cat.

#1 finding squares.

step 1. think of a base which is a multiple of 10 or 100 (whichever nearer to the no. whose square is to be determined.)

suppose 34^2.

a gud base wud be 30.

express 34 as 30 + 4 (i.e. base + 4)...i'll call this base + deviation

step 2.

ldigit(s) before 0 in base x (no. + deviation) | deviation^2 where | is an imaginary line separating 2 parts of calculation.

i.e. 3x(34 + 4) | 4^2

i.e 114 | 16.

now the important thing....no. of digits on right side of the imaginary line shud be exactly same as the no. of zeros in our base..dont forget this!!!!

since our base 30 has only one 0 at the end, v can have only one digit on right side of the imaginary line....the less powerful digit...thats 6.

and thus, one has to be carried to the left side....so 114 will become 115.

so the expression becomes...115 | 6

hence, the answer is 1156.


look at few more examples for practice...

28^2

base 20

2x(28 + 8 ) | 8x8

= 72 | 64

= 784


106^2

106 + 6 | 6x6

base 100

=112 | 36

=11236 . since 100 has 2 zeroes, there shud be 2 digits on right side of the line....


103^2

base 100

103 + 3 | 3x3

=106 | 9

= 10609. ensure the no. of digits are exactly same as the no. of zeroes in base....so expressed 9 as 09....a 2 digit no.

312^2

base 300

3x(312 + 12) | 12^2

3x324 | 144

972 | 144

= 97344....1 of 144 carried.



smart ways of using this method....



98^2. now if v take base 100, v can avoid multiplication by 9.

so base 100.

98 - 2 | -2^2

= 96 | 04

= 9604....funny ha?



197^2

base 200

2x(197 - 3) | -3^2

=388 | 09

= 38809.


#2 : finding product of nos when they lie nearby.


all digits before 0 in base x (first no. + deviation of second no.) | deviation of first no. x deviation of second no.

rest, everything is same...

examples...

27 x 22 =

base 20

2x(27 + 2) | 7x2

= 58 | 14

= 594


103 x 108

base 100

= 103 + 8 | 3x8

= 111 | 24

=11124


97 x 102

base 200

97 + 2 | -3 x 2

=99 | -6

= 9900 - 6

= 9894



197 x 199

base 200

= 2x(197 - 1) | -3 x -1

= 392 | 03

= 39203


37 x 31

base 30

37 + 1) x 3 | 7x1

= 114 | 7

= 1147.


32 x 24

base 30

32 - 8 | 2 x -8

= 24 | -16

= 240 - 16

= 226.


think of few examples, solve them using pen...cross check ur answers...when accuracy becomes gud, try doing it w/o pen....got lotsa other such methods...lemme know if its helpful...if so...wud post more...

hope the post helps...practice guys...practice!!!

regards
maxximus

[iyervani]

Thanks maxximus!!..useful one!!!
btw...eagerly awaiting the posts on remainder thingy also!!!!!

[maxximus]

Originally Posted by iyervani

Thanks maxximus!!..useful one!!!
btw...eagerly awaiting the posts on remainder thingy also!!!!!

rest assured vani....we'll have 3-4 days jus on remainders...soon...n we'll have very-very practical solutions to all the questions....v'll b able to solve 95% of the questions without cramming/theorems...rest 5 % which cannot be solved without a particular theorem do not appear in cat...n wud not...as formula/a theorem based question is neva asked in cat...questions which have a practical...conceptual angle appear...rest assured...u'll tackle them...with lil practice...even without pickin ur pen...jus the way u learnt to find huge products w/o pickin up pen....lil exaggeration...he he he....

wud request everybody around to be ready to pour in their remainders fundas soon...we'll be needin them all...neva know...which approach may click whom...

regards
maxximus

[CAT 2007]

Originally Posted by maxximus

Today's concept....finding squares and close multiplications quickly...

In a hurry guys....so few small concepts that'll help u save sum time...n avoid cramming....which i've always maintained is the best ill-preparation for cat.

#1 finding squares.

step 1. think of a base which is a multiple of 10 or 100 (whichever nearer to the no. whose square is to be determined.)

suppose 34^2.

a gud base wud be 30.

express 34 as 30 + 4 (i.e. base + 4)...i'll call this base + deviation

step 2.

ldigit(s) before 0 in base x (no. + deviation) | deviation^2 where | is an imaginary line separating 2 parts of calculation.

i.e. 3x(30 + 4) | 4^2

i.e 102 | 16.

Just a correction, it should be 3 X (34 + 4) | 16= 114 | 16 = 1156
But a very good way to find out squares... keep rocking man!

[jha ji]

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



@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

maxximus u r doing an incredible and excellent job
i was just going through the first page of this thread and through this problem
as u said to get 3 two's we need to travel 4 places....viz 1x2x3x4
but as we move slightly ahead in the same order we can find 3 two's less frequently
viz...13x14x15x16 i.e the highest power of 2 in this combination is 5i.e 2^5, so as we move further down the order we can combine these surplus two's to form 2^3 and combine them with 3's to get 24
using the greatest integer function method the highest power of 2 in 150! is 146 and the highest power of 3 in 150! is 72
hence the highest power of 24 should be 48 i.e (2^3x3)^48
i think u need not count the places we have to move in order to determine the highest power, if a number whose highest power is to be determined is not prime we can just break it into prime factors and concentrate on the power of the limiting factor i.e factor whose power is less and the conditions with which it can be combined with the other factor
according to me the answer should be 48
please comment and do devote some of your precios time because i am just a novice

[jha ji]

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



@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

maxximus u r doing an incredible and excellent job
i was just going through the first page of this thread and through this problem
as u said to get 3 two's we need to travel 4 places....viz 1x2x3x4
but as we move slightly ahead in the same order we can find 3 two's less frequently
viz...13x14x15x16 i.e the highest power of 2 in this combination is 5i.e 2^5, so as we move further down the order we can combine these surplus two's to form 2^3 and combine them with 3's to get 24
using the greatest integer function method the highest power of 2 in 150! is 146 and the highest power of 3 in 150! is 72
hence the highest power of 24 should be 48 i.e (2^3x3)^48
i think u need not count the places we have to move in order to determine the highest power, if a number whose highest power is to be determined is not prime we can just break it into prime factors and concentrate on the power of the limiting factor i.e factor whose power is less and the conditions with which it can be combined with the other factor
according to me the answer should be 48
please comment and do devote some of your precios time because i am just a novice

[maxximus]

Originally Posted by jha ji

maxximus u r doing an incredible and excellent job
i was just going through the first page of this thread and through this problem
as u said to get 3 two's we need to travel 4 places....viz 1x2x3x4
but as we move slightly ahead in the same order we can find 3 two's less frequently
viz...13x14x15x16 i.e the highest power of 2 in this combination is 5i.e 2^5, so as we move further down the order we can combine these surplus two's to form 2^3 and combine them with 3's to get 24
using the greatest integer function method the highest power of 2 in 150! is 146 and the highest power of 3 in 150! is 72
hence the highest power of 24 should be 48 i.e (2^3x3)^48
i think u need not count the places we have to move in order to determine the highest power, if a number whose highest power is to be determined is not prime we can just break it into prime factors and concentrate on the power of the limiting factor i.e factor whose power is less and the conditions with which it can be combined with the other factor
according to me the answer should be 48
please comment and do devote some of your precios time because i am just a novice

hi jha ji...

thanx for ur appreciation and interest u've shown in the thread...

abt the concept above....plz understand that if u take a particular interval, u may find discrepancies...where did this 13x14x...strike u? dont complicate things....what if i take 25x26x27? now three 3s n only one 2? in case while solving a questn, ur gut feel says otherwise, solve using both prime nos. and one that less wud be the answer..rest assured...u r right...to find answer v dont need to c how many places to travel...but to know which prime no. wud be the d.f, this approach of travellin places does help...but keep in mind...dont pick arbitrarily any set of nos, start from 1...u'll have an idea as to which prime no. is least frequent...in case there is a tie b/w 2 prime nos. for ex. the ques. in which divisor was...144...its difficult to decide whether the d.f. wud be 2 or 3...in such cases, find answers using both..the one thats less wud be the answer...this concept has been discussed over many pages....kindly go thru them...u'll have a fir idea....in case u have any doubts...plz get back...n have faith...nothin discussed here has any fallacy...its abt u grasping it rightly...

24s in 150!

to get 3 2s...4 places

to get one 3...3 places.

but its not easy to say 2 will be the constraint....as higher powers of 2 occur more frequently(4,8,16,32..) than those of 3 (3,9,27,81...) so lets find out answer using both the constraints..

2s... 75 + 37 + 18 + 9 + 4 + 2 + 1 = 146
answer with 2 as constraint = 146/3 = 48

3s... 50 + 16 + 5 + 1 = 72
answer with 3 as constraint = 72

so the answer must be 48.

hope it helps...
feel free to revert...

had it been sth obvious like 99...the constraint has to be 11 (11 places) as 9 (six places) can be easily ruled out...so no need for 2 calculations...jus find no. of 11s.

regards
maxximus

[jha ji]

Originally Posted by maxximus

hi jha ji...

thanx for ur appreciation and interest u've shown in the thread...

abt the concept above....plz understand that if u take a particular interval, u may find discrepancies...where did this 13x14x...strike u? dont complicate things....what if i take 25x26x27? now three 3s n only one 2? in case while solving a questn, ur gut feel says otherwise, solve using both prime nos. and one that less wud be the answer..rest assured...u r right...to find answer v dont need to c how many places to travel...but to know which prime no. wud be the d.f, this approach of travellin places does help...but keep in mind...dont pick arbitrarily any set of nos, start from 1...u'll have an idea as to which prime no. is least frequent...in case there is a tie b/w 2 prime nos. for ex. the ques. in which divisor was...144...its difficult to decide whether the d.f. wud be 2 or 3...in such cases, find answers using both..the one thats less wud be the answer...this concept has been discussed over many pages....kindly go thru them...u'll have a fir idea....in case u have any doubts...plz get back...n have faith...nothin discussed here has any fallacy...its abt u grasping it rightly...

24s in 150!

to get 3 2s...4 places

to get one 3...3 places.

but its not easy to say 2 will be the constraint....as higher powers of 2 occur more frequently(4,8,16,32..) than those of 3 (3,9,27,81...) so lets find out answer using both the constraints..

2s... 75 + 37 + 18 + 9 + 4 + 2 + 1 = 146
answer with 2 as constraint = 146/3 = 48

3s... 50 + 16 + 5 + 1 = 72
answer with 3 as constraint = 72

so the answer must be 48.

hope it helps...
feel free to revert...

had it been sth obvious like 99...the constraint has to be 11 (11 places) as 9 (six places) can be easily ruled out...so no need for 2 calculations...jus find no. of 11s.

regards
maxximus

thanks for ur apt reply

[maxximus]

Today's concept : Finding HCF n LCM of typical values.

#1 : to find the HCF, LCM quickly.

i know most of us wud know this....if v have few nos., 20,40,50,80,180...to find their LCMs, HCF...there's a slightly quick method...

express them in prime nos.

20 = 2^2 x 5
40 = 2^3 x 5
50 = 5^2 x 2
80 = 2^4 x 5
180 =3^2 x 2^2 x 5

now to HCF, see highest power of all prime nos. that are common to all nos.

2 - 2
3- 0
5 - 1

hence hcf is 2^2 x 5 = 20

to find lcm...see highest power of all prime nos across all nos.

2 - 4
3 - 2
5 - 1

hence, lcm = 2^4 x 3^2 x 5 = 1620.


#2 To find HCF and LCM of the form-

2222....30 times.

3333....70 times.

to solve such questions...

for HCF..

take hcf of no. thats being repeated...i.e. hcf of 2 & 3. i.e. 1

take hcf of no. of time these nos. are being repeated...i.e. hcf of 30 n 70...thats 10.

so the hcf is 111...written 10 times.

For LCM...

take lcm of no. thats being repeated...i.e. lcm of 2 & 3. i.e. 6

take lcm of no. of time these nos. are being repeated...i.e. lcm of 30 n 70...thats 210.

so the hcf is 666...written 210 times.

#3...to find hcf and lcm of following form...

2^300 - 1, 8^250 - 1.

the idea is..a^n - b^n is always divisible by a-b. so v need to find highest a-b that will divide a^n - b^n and smallest term that'll be divisible by a^n - b^n.

express them in a common base.

2^300 - 1 and 2^750 -1.

to find hcf...

take hcf of powers i.e. hcf of 300 and 750...i.e. 150

so the hcf is 2^150 - 1.

to find lcm....

take lcm of powers i.e. lcm of 300 and 750...i.e. 1500

so the hcf is 2^1500 - 1.




Questions :

find hcf and lcm of:

1.2222...250 times and 8888...300 times

2. 333....120 times and 1111...400 times

3. 111...700 times and 9999...200 times.

4. HCF of 33333...200 times. and 777777.....300 times

5. 32^250 -1 & 16 ^ 100 - 1.

6. 81^100 -1 & 243 ^ 200 - 1.

7. 343^150-1 & 2401^100 - 1.

8. 125^200 - 1 & 625^120 - 1.

9. 169^320 - 1 & 32^160 - 1.

for the following questns...

mark 1. - stmt 1 is sufficient.
2- stmt 2 is suff.
3-both are reqd to solve the questn.
4-either is suff.
5-both insufficient.

10. what is the hcf of 5 nos., a,b,c,d,e?

stmt 1 - a=72,b=4,c=6
stmt 2 - d= 8, e = 27.

easy set...hope many wud get all correct...

regards
maxximus