[akash_b]

Originally Posted by thakarsagar

yes...suresh, my name is sagar...



Am tired of remainder problems...here is a different one...

i have infinite no of 1 Re, 5 Re, 10 Re and 25 Re notes...in how many ways can i give the change for Rs 100???
-sagar

joined pagal guy yesterday..and it gives me great satsfaction to contribute in this thread

acc. to me the ans is 242
the method is lil prehistoric ...but nonetheless the prob can be solved under 2 mins


soln a + 5b + 10c + 25d = 100
5b + 10c + 25d <= 100
here we try to get all the possible comb. so that the addition is =< 100. the add. if below 100 will be automatically adjusted by diff vales of a
for d = 0; c= 0,1,2....10
for c =0 ; d= 0,1,...20 ( 21 values )
slly, for c=1; d= 0,1....18 ( 19 values )
so on till c =10; d=0 ( 1 values )

similarly for d=1,2,3,4
the key is that since 10 is a multiple of 5 ,hence the total poss val are in ap , u need not count & addall the values. u can simply take the avg and multiply with the count
otherwise its tough to solve these kind of sums under time constrain

[akash_b]

saw the above prob solved in the later pages
anyways would appreciate if someone could suggest total possible soln for smthing like
2x + 3y + 25z = 500

[anupam_khanna]

Originally Posted by clsuresh

Folks, I am back again into my thread. I will be continously posting questions from now till the CAT exam( as done by me in the last year). Infact I'll be concentrating a lot on numbers, PG and pure maths part.

Here comes our first question.
1. Remainder when (56)^493 is divided by 126.

breaking the denominator into least for we get

f(x)= 4 * 56^492 / 9

now 56 ^ 492 mod 9 = 1

therefore f(x) = 4 * 56^492 mod 9
=> 4

now sice we divided by 14 to make the num and den coprime we multiply this by 14 to yeild the final remainder...
therefore final rem = 4 * 14 = 56

cheers!!!

[ak_shri250]

Originally Posted by clsuresh

Thakar,

Well done, Can u please tell me the way in which u answered. ( did u use FERMAT's rule)

Help me friends, What is FERMAT's rule???
ashu

[ninja_talli]

Originally Posted by ak_shri250

Help me friends, What is FERMAT's rule???
ashu

well i'll just state fermat's theorem..

for two co-prime nos a & p,

a^p = a(mod p)
or in other words
a^(p-1) = 1 (mod p)

[Pragati Soni]

Hello navi,

Would u plz make it clear....if u r saying probability 1/8(exact)..then does it imply.....out of 8 methods we can break the stick only 1 method is their,,in which we can form the Triangle...

I am clear with the explaination that 1/2<c<1/2 but I am not clear with d probablity....

Plz tell me.

thanks

Originally Posted by nagi_nov3

u have a valid point

apart from a,b<1/2
we will have a+b > 1-(a+b) ( sum of 2 is more that 3rd one)
a+b >1/2
so third side 1-(a+b)/2 also <1/2.

so answer is less than 1/8.

Thanks buddy.

[prade]

Originally Posted by nagi_nov3

u have a valid point

apart from a,b<1/2
we will have a+b > 1-(a+b) ( sum of 2 is more that 3rd one)
a+b >1/2
so third side 1-(a+b)/2 also <1/2.

so answer is less than 1/8.

Thanks buddy.

let the two cuts be made of lengths a and b from one end.

let b>a
hence the three sides are a, b-a, 1-b

now a+b-a > 1-b => b>1/2
b-a+1-b> a => a<1/2
1-b+a > b-a => b<a+1/2

and plot the above on the x and y axis we get a square of side 1. and the common region satisfying above eqns is 1/8 of the square.

again when we consider a>b we get 1/8 as the common area

hence the probability = 1/8 + 1/8 = 1/4...

[nagi_nov3]

Originally Posted by prade

let the two cuts be made of lengths a and b from one end.

let b>a
hence the three sides are a, b-a, 1-b

now a+b-a > 1-b => b>1/2
b-a+1-b> a => a<1/2
1-b+a > b-a => b<a+1/2

and plot the above on the x and y axis we get a square of side 1. and the common region satisfying above eqns is 1/8 of the square.

again when we consider a>b we get 1/8 as the common area

hence the probability = 1/8 + 1/8 = 1/4...

i think u shud not consider a>b
beacuse

in the first case u have considered a<b (2/7 <3/7)
next case u dont need to consider a>b (3/7 > 2/7) ( please try to prove this wrong with e.g.)

what ever the case, it might have covered in one case.

so i think answer is 1/8.

[prade]

Originally Posted by nagi_nov3

i think u shud not consider a>b
beacuse

in the first case u have considered a<b (2/7 <3/7)
next case u dont need to consider a>b (3/7 > 2/7) ( please try to prove this wrong with e.g.)

what ever the case, it might have covered in one case.

so i think answer is 1/8.

try to plot the graph u will get it.. will post wit some expln later...

[nebuchadnezzar]

hi clsuresh/others...

help me with this one...

A rectangular floor is fully covered with square tiles of identical size. The tiles on the edges are white and the tiles in the interior are red. The number of white tiles is the same as the number of red tiles.
A possible value of the number of tiles along one edge of the floor is:

1. 10 2. 12 3. 14 4. 16

lemme tell my approach...

i take the value of the sides as n1 and n2 ..then (n1-1)(n2-1) = n1*n2 - (n1-1)(n2-1),

or, 2(n1-1)(n2-1) = n1* n2

for any of the answer options, the eqn is not giving an integer value for the other number...what is wrong with my approach..???

Vaibhav