[shashank3012]

Puys,its MaskedMenace's bday today.wish him here

http://www.pagalguy.com/forum/pagalg...ml#post1351488 (Happy Birthday MaskedMenace (16th Jan 2009))

for those who don't know,you will soon be finding out who he is in the coming days

[rahulshaitan]

from my side

divisibility rules




Very often while solving questions the divisibility rules prove to be very handy and we can save a lot of time while solving the questions.
Some of the divisibility rules are as follows-


Divisibility by 2,4,8,16,….

A number is divisible by 2,4,8,16,….,2n when the number formed by the last one, two, three, four, …., n digits is divisible by 2,4,8,16,….,2n respectively.





Divisibility by 3



A number is divisible by three if the sum of its digits is divisible by 3.




Divisibility by 5


A number is divisible by 5 if the digit at unit’s place is 0 or 5.



Divisibility test of 7


Whenever we have to check whether a number is divisible by 7 or not follow the below procedure:



1).Double the last digit (digit at the rightmost place) and subtract it from the number left (excluding the last digit). If this number is divisible with 7 then the original number is divisible by 7.


This procedure can be followed as many times as required (until the number is reduced to 2 digit number). Then the number so obtained can be checked whether it is divisible by 7 or not. If the number so obtained is divisible by 7 then the original number is divisible by 7 and if not then original number is not divisible by 7.


E.g.-
Consider the number 1057.
Now the last digit is 7. On doubling it we get 14.
On subtracting it from 105 we get 91.
Now it can be seen that 91 is divisible by 7 so the original number is divisible by 7.


(It can further be simplified by doubling 1 and subtracting it from 9 and thus we get 7 which is divisible by7.)


Divisibility test of 8


To check whether a number is divisible by 8 or not we consider the last three digits of the number. If the last three digits are divisible by 8 then the whole number is divisible by 8 otherwise not.
E.g.-7645892
Now consider the last three digits 892. Now check whether the three digits are divisible by 8. On checking we get that the number is not divisible by 8 so the original number is not divisible by 8.



Divisibility by 9

A number is divisible by three if the sum of its digits is divisible by 9.


Divisibility test of 10

If last digit of a number is 0 then the given number is divisible by 10 otherwise not.
E.g.- 9860 is divisible by 10 as the last digit is 0.

divisibility test for 11

If the difference between sum of digits in the odd places and the sum of the digits in the even places is either 0 or is divisible by 11, then the number is divisible by 11.



Divisibility test of 13
To check whether a number is divisible by 13 we follow the procedure as follows:
1). Multiply the last digit with 4 and add it to the number left (after removing the last digit).
2). Follow this method again and again and reduce the number to 2-digit number form.
3). Now check whether the number is divisible by 13 or not.
If the 2-digit number so obtained is divisible by 13 then the original number is divisible by 13 otherwise not.
E.g.-
Let us consider the number 195.
Now the last digit is 5 and on multiplying it with 4 we get 20.
Now on adding this with the remaining number (i.e. 19) we get 39. Now as 39 is divisible by 13 therefore the original number id divisible by 13.

Divisibility test of 17
To check whether a number is divisible by 17 we follow the procedure as follows:
1). Multiply the last digit with 5 and subtract it from the number left (after removing the last digit).
2). Follow this method again and again and reduce the number to 2-digit number form.
3). Now check whether the number is divisible by 17 or not.
If the 2-digit number so obtained is divisible by 17 then the original number is divisible by 17 otherwise not.
E.g.-
Let us consider the number 221.
Now the last digit is 1 and on multiplying it with 5 we get 5.
Now on subtracting 5 from the remaining number (i.e. 22) we get 17. Now as 17 is divisible by 17 therefore the original number id divisible by 17.


Divisibility test of 19


To check whether a number is divisible by 19 we follow the procedure as follows:
1). Multiply the last digit with 2 and add it to the number left (after removing the last digit).
2). Follow this method again and again and reduce the number to 2-digit number form.
3). Now check whether the number is divisible by 19 or not.
If the 2-digit number so obtained is divisible by 19 then the original number is divisible by 19 otherwise not.
E.g.-
Let us consider the number 209.
Now the last digit is 9 and on multiplying it with 2 we get 18.
Now on adding th18 with the remaining number(i.e. 20) we get 38. Now as 38 is divisible by 19 therefore the original number id divisible by 19.


Divisibility by 7, 11, 13
Consider any number abcdefghij.
Starting from the right towards left, we make groups of 3 digit numbers successively and continue till the end. It is not necessary that the leftmost group has three digits.
Now we have
a bcd efg hij
Now we number these groups starting from the left.
So, we have
Group 1= hij
Group 2= efg
Group 3= bcd
Group 4= a
And if the number is bigger this number can be continued.
Now, we add the alternate groups i.e. 1,3,5,… and 2,4,6,…
So we get two sums
Sum1= group(1+3+5+….)
And
Sum2=group(2+4+6+….)
Now we find the difference between the two sums obtained i.e. sum1 and sum2.
Let the difference be D.
D= sum1 - sum2
Now if D is-
i). divisible by 7, then the original number is divisible by 7.
ii). divisible by 11, then the original number is divisible by 11.
iii). divisible by 13, then the original number is divisible by 13.


Corollary: Any 6-digit or 12-digit ,or any such number with number of digits equal to multiple of 6,is divisible by each of 7,11 and 13 if all of its digits are same. E.g. 666666, 888888888888 etc. are divisible by 7,11 and 13.




Divisibility rules for composite numbers like 6,12,14,18,….


Whenever we have to check the divisibility of a number N by a composite number C, then the number N should be divisible by all the prime factors (the highest power of every prime factor) present in c.
Now the divisibility rules for various composite numbers will be-




1). Divisibility by 6




The given number should be divisible by both 2 and 3.




2). Divisibility by 12





The given number should be divisible by both 4 and 3.




3).Divisibility by 14




The given number should be divisible by both 2 and 7.




4). Divisibility by 15





The given number should be divisible by both 3 and 5.




5). Divisibility by 18



The given number should be divisible by both 2 and 9.




happy learning

[navneet023]

Puys...puys puys....

A request to all. I think we are going too fast with concepts....problem solving is ok. But let's take some time to absorb the conceptual thing. I think one long post on concepts is more than enough for 2 days. We should wait for everyone to get into discussion. Its not like CAT is nxt month or something like that. Problem solving is ok and even for that wait till atleast 4-5 ppl post answers and explanations. This ainit no exam. Here we have to build up the concepts. So I request everyone to go a bit slow.

Cheers,
Navi

[astute10]

Originally Posted by shashank3012

1)x!=1.25(y!)^2
2)504=2*2*2*3*3*7

as 7 is the least frequently occurring,we have to find number of 7's in 10200
10200/7=1457
10200/49=208
10200/343=29
10200/2401=4

total of 1698

so,n=1698.

Hi,

Can u explain why 7 is the least occurring..............if I see with powers then 3^2 = 9 shud be the least occurring

[astute10]

Originally Posted by srebalajii

Soln:

1.
From the Q: U can frame the Eq lik thiz x!/(y!)^2

Then, x!/(y!)^2 is increased by 50% ==> 1.25
Thfr: (3/2)*x!/(y!)^2 = 5/4
By solving: u get (y!)^2 = (6/5)*x!

Now, Apply ARM to x! == If u get a perfect sq val then it must be the Answer.

Hence for opt (b) 5 [We are getting a perfect sq value for (y!^2 == 144)] == Answer

From the condition we know y = 2/3 X .............. Now there seems to be two errors ............1) 2/3(x!) is not same as (2/3X)! eg : X=3 ; 2/3(x!)=4 and (2/3X)! =2 ...SO u can't say X/ ((2/3X)!)^2 is same as 9/4 X /X!^2 ..........2) expl of second is in the first......its for u to find

[celebnavin]

A perecious stone worth Rs 10872 fell and broke into three pieces , the wieghts of which are proportional to 1:2:3 . The values of each stone is directly prop. to the square of its weight. Find the loss in value caused due to breakage.

1) 3624 2)4228 3)6644 4)7510

Can sum1 help me with this question.

[slam]

Originally Posted by celebnavin

A perecious stone worth Rs 10872 fell and broke into three pieces , the wieghts of which are proportional to 1:2:3 . The values of each stone is directly prop. to the square of its weight. Find the loss in value caused due to breakage.

1) 3624 2)4228 3)6644 4)7510

Can sum1 help me with this question.

given that value = K*m^2, where K = constant, m = weight of the stone

initially,
10872 = K (6x)^2 , taking the initial weight to be 6x
=> Kx^2 = 10872/36

after breaking the stone
total value = K(3x)^2 + K(2x)^2 + K(x)^2
= 14Kx^2
= 14 * 10872 / 36 = 4228

=> difference = 10872 - 4228 = 6644. Is this correct?

[celebnavin]

Originally Posted by slam

given that value = K*m^2, where K = constant, m = weight of the stone

initially,
10872 = K (6x)^2 , taking the initial weight to be 6x
=> Kx^2 = 10872/36

after breaking the stone
total value = K(3x)^2 + K(2x)^2 + K(x)^2
= 14Kx^2
= 14 * 10872 / 36 = 4228

=> difference = 10872 - 4228 = 6644. Is this correct?

yeah , tht's ryt.. thanks slam !!

[Vshrimenon]

Originally Posted by rahulshaitan

from my side

divisibility rules









divisibility test for 11

If the difference between sum of digits in the odd places and the sum of the digits in the even places is either 0 or is divisible by 11, then the number is divisible by 11.

Hey since u told about the divisibilty of 11 i have an easy method for the finding the product of any number with 11

say u have a 3 digit num. say abc and u need 2 find

abc * 11 = a _ _ c

Retain the first and the last digit as a and c

then sum up b and c , u get the ten's place

abc * 11 = a _ (b+c) c

Now sum up a and b , u get the hundred's place

abc * 11 = a (a+b) (b+c) c

eg: 172 * 11 = 1 _ _ 2

sum up 7 and 2 = 9

so 172 * 11 = 1 _ 9 2

sum up 1 and 7 = 8

so 172 * 11 = 1 8 9 2

this is true for any N digit u want to multiuply with 11

U just need 2 folow 2 things

1) Apply the rule from right to left

2) In case u get the sum 2 be 10 or getter than that then carry over to the next digit addition

say 146 * 11 = 1 6 0 6

since 4+6 = 10
u will carry the 1 when adding 1and 4 ( for the 100th place)

Thought it wil be useful

[Aarav]

The official thread of 2008 was very useful for the aspirants - and 2009 thread is likely to be on similar lines.By the time we come to CAT 2009, this thread will run into 1000 pages and hence I will puys need to evolve a method where questions and concepts get more structured here. May be using a Set concept (month-wise) and numbering each question serially (will help in refer backs).

I'm posing a question that to my knowledge has a concept that needs to be evolved and it requires little background to existing mathematical concepts. This will help develop god line of thought.


Let S be a subset of {1, 2, 3, ... , 10} such that no two subsets of S have the same sum. What is the largest possible sum for S?

(1) 28 (2) 30 (3) 31 (4) 33 (5) none of these


Let S be a subset of {1, 2, 3, ... , 15} such that no two subsets of S have the same sum. What is the largest possible sum for S?

(1) 55 (2) 58 (3) 61 (4) 63 (5) none of these


Let S be a subset of {1, 2, 3, ... , 30} such that no two subsets of S have the same sum. What is the largest possible sum for S?

(1) 150 (2) 153 (3) 161 (4) 162 (5) none of these