please help ..

[Achillies]

1 If (x+1/x) / (1+1/x) = 99 then Find value of (x-1/x)/(1+1/x) = ?
2 When 10^ (22) + 1 is divided by 11, what is the reminder?

[chestnut.cc]

1.Is that (x+1/x) or ((x+1)/x)...

2.When 10^n is such that n is even the remainder is 1... If n is odd... the remainder is 10... hence the ans is 2 since n is even...

[Achillies]

Thanx chestnut1.Is that (x+1/x) or ((x+1)/x)...it is (x+1/x)

[julle286]

Originally Posted by chestnut.cc

2.When 10^n is such that n is even the remainder is 1... If n is odd... the remainder is 10... hence the ans is 2 since n is even...

Hi Chestnut,

why that?

[julle286]

Question A: Soltution is 98.

(x+1)/x / (1+1)/x =99

--> (x+1)/2 = 99

--> x = 197


IF you plug it in you get (196/197) / (2/197) , which is 98.....


Should be this way ;-)

[Achillies]

@julle286
actually its(x+(1/x)) / (1+(1/x)) = 99 then Find value of (x-(1/x))/(1+(1/x)) = ?
expln for prob 2
=rem((10^22+1)/11)
=rem(10^22/11) + rem(1/11)=1+1.......since u tak any thing greater than 11 i.e 10^10=100 div by 11 will give remainder as 1
=2
hey chestnut plz help with prob 1

[chestnut.cc]

Thats what i asked him... if it is (x+1)/x its simple... not so much if its the latter...

If you try out a few random odd and even powers of 10... youll see that 11 will neatly divide upto 9999...n for every even power... and 9999...0... for every odd power...

Maybe this could be a rigorous exposition... :s

Let n be even...

10^n/11

Split up as 100*10^(n-2)

= 100*10^(n-2)/11
= ((99+1)/11)*10^(n-2)
= 99*10^(n-2)/11 + 1*10^(n-2)/11
= 9*10^(n-2) + 10^(n-2)/11

Now do the same for 10^(n-2)

= 9*10^(n-2) + 99*10^(n-4)/11 + 1*10^(n-4)/11
= 9*10^(n-2) + 9*10^(n-4) + 10^(n-4)/11

Now since n is even the last term in the generalized series is 10^(n-n)... and the N in the second term is 1 since it is 10^(n-n)... ie 10^0 =1...

Simplify and extend to the general case when n = even as follows...

=9*10^n[ 1/10^2 + 1/10^4 + .... + 1/10^n] + 1/11

Lets test our generalization... Let n=6... plug it into the formula...

9*10^6[1/10^2 + 1/10^4 +1/10^6] + 1/11
= 9*10^4 + 9^10^2 + 9 + 1/11
= 90000 + 900 + 9 + 1/11
= 90909 + 1/11
=(999999 +1)/11
=1000000/11
=(10^6)/11



Which is indeed the expression for (10^6)/11...

What if n is odd... in that case the last term in the first bracket will be 1/10^m where m is the least even no before n... and the N in the second term will be 10^(n-m)/11...

Lets test it for n=3

=9*10^3[1/10^2] + 10^(3-2)/11
=9000(1/100) + 10/11
=90 + 10/11
=990+10/11
=(10^3)/11

Which is indeed the expression for (10^3)/11... you can see the remainder is the second term... which is 10 when n is odd... and 1 when n is even...

[chestnut.cc]

(x+(1/x)) / (1+(1/x)) = 99

Let y be the common factor...

(x^2 + 1)/(x+1) = 99y/y

x^2 +1 = 99y
x+1 =y

(x+1)(x-1) = 99y -2
(x-1) = 99 - 2/y

(x-(1/x))/(1+(1/x))
(x-1)(x+1)/(x+1)
(x-1)
99 - 2/y

How you find y... i dont have a clue... you would have to look at the ans and guess i sup...

[Achillies]

hmm..may be...actually i came across one link which has the recent gre questions ....and there ans are not given..anyways....thanx

[chestnut.cc]

read what hap to scoretop.com before doing that... peace...