[ashish banaya..]

Originally Posted by amar_kashyap

problem#132
Let . Find the smallest such that divides

Is the ans. 1059
Reason:2007=223*9.2007^2007=(223^2007)*(9^2007)
9^2007 can be taken care of.
As regards 223^2007,the 1st 223 will be divided in the 223;(let ; be factorial). And this will continue upto 445;(1*223)
From the 446; there will be two 223's in every ; upto 668;(2*223)
From the 669; there will be three 223's in every ; upto 891;(3*223)
And from 892; there will be four 223's in every ;upto the ans.
So the smallest X=891;+168(i.e. remaining 2007-6*223/4)
Therefore the ans. is 1059

Disclaimer:This is what i think,cud be wrong

[amritonline]

Originally Posted by prahalad85

yeah Even i am getting 2229.
reason being
2007=9*223
thus 223^2007 WE NEED AN X VALUE OF ATLEAST 223+2007-1=2229

Me also getting answer 2229. Is it correct ??

[amar_kashyap]

Originally Posted by ashish banaya..

Is the ans. 1059
Reason:2007=223*9.2007^2007=(223^2007)*(9^2007)
9^2007 can be taken care of.
As regards 223^2007,the 1st 223 will be divided in the 223;(let ; be factorial). And this will continue upto 445;(1*223)
From the 446; there will be two 223's in every ; upto 668;(2*223)
From the 669; there will be three 223's in every ; upto 891;(3*223)
And from 892; there will be four 223's in every ;upto the ans.
So the smallest X=891;+168(i.e. remaining 2007-6*223/4)
Therefore the ans. is 1059

Disclaimer:This is what i think,cud be wrong

Me too getting the same..
The approach is perfect...

[amar_kashyap]

problem #133
Vaibhav wrote a certain number of positive prime numbers on a piece of paper. Vikram wrote down the product of all the possible triplets among those numbers. For every pair of numbers written by Vikram, Vishal wrote down the corresponding GCD. If 90 of the
numbers written by Vishal were prime, how many numbers did Vaibhav write?

(1)6 (2)8 (3)10 (4) Cannot be determined

[warrior]

Originally Posted by amar_kashyap

problem#132
Let . Find the smallest such that divides

me solved it like this 2007=223*9 (223 is prime)

now 223! and above has 223! in them. Also 223^2is a large no so no repitation can occur.

so no terms we need is 223+2007-1=2229

also 3 will occur so many time that it will take care of 9^2007 = 3^6021

[varun nakra1]

Originally Posted by amar_kashyap

problem #133
Vaibhav wrote a certain number of positive prime numbers on a piece of paper. Vikram wrote down the product of all the possible triplets among those numbers. For every pair of numbers written by Vikram, Vishal wrote down the corresponding GCD. If 90 of the
numbers written by Vishal were prime, how many numbers did Vaibhav write?

(1)6 (2)8 (3)10 (4) Cannot be determined

Hi dost.kaisa hai-
Is the answer 8??Me gettin 8 by (8/*[C(6,2)*C(4,2)]=90..tu bata if the answer is rite then ill post the method

[sandeep_slr]

Originally Posted by sanket

consider that being at same end is not meeting in the course...it cant be 6. if its 4m/s they r at the same end together 2 times out of a total of 19 times that they can meet i.e at 100 secs and 200 secs..therefore they meet 17 times.

4 m/s....is it rite???

[amar_kashyap]

Originally Posted by varun nakra1

Hi dost.kaisa hai-
Is the answer 8??Me gettin 8 by (8/*[C(6,2)*C(4,2)]=90..tu bata if the answer is rite then ill post the method

I am getting 6 as answer.
6. For total number of primes at the end= 6.(5C2.3C2)/2=90.

Plz..somebody check...

[amar_kashyap]

problem#134
If

find ..

[warrior]

Originally Posted by amar_kashyap

problem#134
If

find ..

m= 64

just multiply and divide LHS by (1-1/x^2) and simplyfy