[sanket]

Originally Posted by ashish banaya..

Question 131#

A and B start simultaneously at one end of a swimming pool whose length is 50 m. The swimming race is a race of 1000 m. If A beats B and meets him 17 times during the course. If A’s speed is 5 m/s then the speed of B could be

1. 1 m/s 2. 3 m/s 3. 4 m/s 4. 6 m/s

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.

["DHRUV"]

Originally Posted by rajat_nda

i suppose the property used is :
a=b mod(m) & c=d mod(m)
then
ac=bd mod(m).

so wht i can assimilaate is tht u write

13^80 = 1 mod(m)
13^2 * 13^78 = 1 mod(m)
13^2 * 13^78 = 169*y mod (m)
-----------------= 69*y mod m
---------------- = 69 *29 mod(m)
but howw did u get 29 ?
should we keep tryin...69*9...69*19...69*29.....andd so on for other
questions or do u have a better one

plzz enlighten my sole
ciao

no hit n trial yaar .....

u have to get 1 so 9 shund be multiplied with a number with last digit as 9 , now a carry of 8 will be taken , , now sine uir require d number is efectively 001 after multiplication , so in order to get that , the carry 8 shud be added to some tnumbe ending with 2 which will only be 72 \which is 9*8..
hence 2nd digit will be 8 , so number will be 89 ..

if u fail to get it ..plz pm me ...

[rajat_nda]

Originally Posted by DHRUV

no hit n trial yaar .....

u have to get 1 so 9 shund be multiplied with a number with last digit as 9 , now a carry of 8 will be taken , , now sine uir require d number is efectively 001 after multiplication , so in order to get that , the carry 8 shud be added to some tnumbe ending with 2 which will only be 72 \which is 9*8..
hence 2nd digit will be 8 , so number will be 89 ..

if u fail to get it ..plz pm me ...

spiffin....ripping n words redolent to d same.....
things seem so obvious wen they r in black n white.

thanks dude...crystal clear

[amar_kashyap]

problem#132
Let . Find the smallest such that divides

[warrior]

Originally Posted by amar_kashyap

problem#132
Let . Find the smallest such that divides

is the answer 2229

[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

[rajat_nda]

@prahlad(still not following why 2007+229-1,why not 229)
and
@warrior,

kindly bother urself 2
explicate ur respctve apprchz plz

ciao

[amar_kashyap]

Originally Posted by warrior

is the answer 2229

I don't know the answer brother..!!.
But I guess...the answer..is not right...
It will be great if you guys post the approach..

[prahalad85]

well the reason for adding 2007 is this...
given that .
hence v need 223 to repeat atleast 2007 times.so we have to increase the value of 223 by further 2007.
this makes the x value divisible .well i think i made a mistake in that -1.
my answer is 2230
i hope i am clear..

[richa bhatia]

Originally Posted by amar_kashyap

The interval in option a) includes the interval in option c) also...How do we check for that..
I think there is more to it. We can proceed by taking into account the fact that if 2<x<3 (according to the question, there is one solution between 2 and 3), then the quadratic equation in x-[x] will have a solution in the interval 0<x-[x] <1.That means F(x-[x]) will have one solution in the interval (0, 1). That implies F(0).F(1)<0 , tht implies a^2.(a^2+2+a-2)<0, tht implies a(a+1)<0. Hence -1<a<0. Hope i make sense .

x-[x] always lies in the inerval (0,1) i.e. 0<x-[x]<1
so both the roots of equation F(x-[x]) should lie in the interval (0,1)
so we can not say that F(0).F(1)<0
rather as the number of roots is even so F(0).F(1)>0

Originally Posted by Aarav

(a-2)(x-2)^2 + 2(x-2) + a^2 = 0
=> 4 -4(a-2)a^2 >= 0
=> 1 - a^3 - 2a^2 >= 0
=> (1+a) (a^2 + a -1) <= 0
=> a lies in (-1, (5^1/2 -1)/2)

4-4(a-2)a^2>=0
=>1-a^3+2a^2>=0

A root of the eq lies in the interval(2,3)
so (a-2)(x-[x])^2 +2(x-[x]) +a^2 =0
=> (a-2)(x-2)^2 + 2 (x-2) +a^2 =0

As there is only unique solution in the interval (2,3) so for above equation D should be 0
=> a^3-2a^2 -1=0

the rate of growth of a^3>a^2
but the growth of 2a^2 > a^3 till 2
after tht a^3 will take over
so a should be >2

a should be in the interval (2,3)

mistakes wud have been committed please rectify :neutral: