Number System

[nakul05]

Originally Posted by Varun Khullar

can somebody give me the answer to

32^32^... till infinity.. divided by. 9.. plz..

u mean to find remainder or the quotient//??
But in any case, is it possible to find it with infinity in numerator?

[chetna]

--edited--

[rajat_nda]

Originally Posted by Varun Khullar

@chetna and the other guys..

think in terms of inverse..
i agree to remainder theorem etc..

we understand.. 31*4 =124 =123+1..

so .. we got.. 31*4 =1mod123
now.. we got 31^-2 which is .. same as 4^2 =16..

seems to b viable dude...
plz tell me....how come 31^-2 b same as 4^2,i'm lost...
anybody...humour me....

[chetna]

Originally Posted by rajat_nda

seems to b viable dude...
plz tell me....how come 31^-2 b same as 4^2,i'm lost...
anybody...humour me....

hey rajat ,
here goes the problem:-

31^78/123 find the remainder

..
so 31*4=124=123+1....its clear..i know

31^78=31^80*31^-2
thus,

31^78=31^-2 mod123
we can write it as:-
=31^-2 * 4^-2 *4^2 mod123
=(31*4)^-2*4^2 mod 123
=124^-2 * 4^2 mod123
=1*4^2mod123
=16mod123

i hope u got it now...
goodnite

p.s-rajat,ur inbox limit has exceeded..plz clear unrequired ones...i m pasting the solution here

[dushyant bhatia]

Originally Posted by chetna

wow...great answer varun....good one

i don't think this is the rite way to do it.. coz sumtimes the answer is incorrect

take for eg.. reamainder when 8^828 is divided by 167
the answer is 107.. but if we follow varun's method we get 64 which is incorrect..
correct me if Im wrong
cheerz

[Varun Khullar]

Originally Posted by dushyant bhatia

i don't think this is the rite way to do it.. coz sumtimes the answer is incorrect

take for eg.. reamainder when 8^828 is divided by 167
the answer is 107.. but if we follow varun's method we get 64 which is incorrect..
correct me if Im wrong
cheerz

could u explain .. how u get 64 dude..
u get 441.. and dividing by 167.. u get 107..

[dushyant bhatia]

Originally Posted by Varun Khullar

could u explain .. how u get 64 dude..
u get 441.. and dividing by 167.. u get 107..

sry broda my mistake..i did sumthin else.. neways cud u just post the solution to this question..

[komal.m]

Originally Posted by dushyant bhatia

my approach wud be as follows:
45 = 5*9
when divided by 9 it gives no remainder
(as the sum of its digits is divisible by 9 ..i.e.(44+1)+(43+2)+(42+3)......(23+22))
Remainder when the no is divided by 5 = 4(as can be seen.. last 2 digits 44)

hence the remainder is 4..
cheerz

ans is 9
even i got 4 bt i dnt know hw ans can b 9.
well thanx all of u for ur response
ill wait 4 the ans

[komal.m]

Originally Posted by clsuresh

Dear friend ur mistaken.

U said that the remainder obtained when this number is divided by 9 is 0.
Now since 9 is a factor of 45 the remainder obtained when the give number is divided by 45 should be such that it should leave remiander 0 when divided by 9. Hence from the options the answer must be either 9 or 18.

Now let's take the other factor of 45 i.e 5. It's clear that the given number leaves a remainder 4 when divided by 5. Hence the remainder must be such that when i divide it by 5 i should get remainder 4.
So clearly out of that 9 and 18, 9 must be the answer.
Hence the answer is 9.


Guys, I would like all of u to go through this basic funda for more clarity

Let R be the remainder when N is divided by D1.
Now the remainder when N is divided by D2, where D2 is a factor of D1 is either
1. R ( if R< D2) or
2. R1 where R1 is the ramainder obtained when R is divided by D2 ( if R>D2)
Remember the converse is also true.

Let's look at an example here. ( let me give the converse example)

Let 23 be the remainder obtained when a number is N divded by 30.
Now the remainder obtained when the number is divided 90 could be either
23 or 30x1+23 or 30x2+23 i.e
23 or 53 or 83. ( I think this is clear to everyone)

Applying this rules to the following queries

1. last two digits of N = 2^999.
All of us know that the remainder obtained when a number is
a) divided by 10 the remainder is the unit digit
b) divided by 100 the remainder is the last two digits of the number
c) divided by 1000 the remainder is the last four digits of the number and so on.

So let's divide 2^ 999 with 100 and find the ramainder.
First find the remainder obtained when it is divided by 25 ( since 25 is factor of 100).

2^999 % 25
= (1024^99) x 2^9 % 25
Remainder obtained when 1024 is divided by 25 is 24 but we can take it as -1 also( always take the smallest remainder irrespective of sign)
So finally the problem becomes
-1 x 2^9 %25
= -512 %25
= -12 =13 ( finally give the positive remainder)
So when our number is divided by 25 the remainder is 13
So when it is divided by 100 the remainders could be
13 or 38 or 63 or 88.
It is evident that the given number is exactly divisible 4.
I am taking 4 here because the other factor of 100 is 4.
So from the above three remainders there is only one remainder which is exactly divisible by 4 i.e 88.
So the remainder obtained when it is divided by 100 is 88 and thus the last two digits are 88.
Remember here once I decide the remainders using 25 in the exam I can make use of the multiple options and then proceed.
So folks remember when the divisor is big make use of its factors and options given in the exam

2.
Remainder obtained when 2^1990 is divided by 1990.
Again the factor here are 10 and 199
We are lucky here since 199 is prime and let's use our FERMAT'S rule.

So 2^198 when divided by 199 the remainder is 1( Fermat's rule).
So 2^199 will leave a remainder 2
So 2^ 1990
= (2^199)^10 % 199
= (2^10) % 199 = 29.
Now when 2^1990 is divided by 10 the remainder is nothing but the unit's digit and we know we can calculat unit's digit by using the cycylicity of 2( i.e after every 2^4 the unit's digit will repeat.) So obviously the unit's digit is nothing but 2^2 =4.

Now since we got the remainder as 29 when divided by 199
the remainders possible when divided by 1990 are
29 or 199x1+29 or 199x2+29.........
But since the unit's digit is 4 the only possible remainder is
199x5+ 29 =1024

That's it folks. These two examples tell u how powerfull our rule is. I think from now however big the divisor is I think we should be able to give the remainders because If the divisor is big and prime please phone a friend (i.e FERMAT) if it's not prime don't call me please make use of the factors of the divisors and ur options in the exam.

I think I am clear.

For more information please go through my thread name Suresh's Corner:
http://www.pagalguy.com/forum/quanti...ns-quants.html (Suresh's Corner: Miscellaneous Questions From Quants)

Regards,
Suresh.

thanx suresh

[rajat_nda]

Originally Posted by Varun Khullar

could u explain .. how u get 64 dude..
u get 441.. and dividing by 167.. u get 107..

On @varun's lines...
v have 8^828 by 167. now ____^166 = 1 mod 167
ergo,8^830=1 mod 167
or we have 8^-2 =x mod 167
Now we take 21^2 and 21^-2 (juz divide 167 by 8(whole num) n add 1 to get dis magic number)
so 8^-2 * 21^-2 * 21^2 = x mod 167,
or, 441 = x mod 167
x = 107

@ komal.m plz avoid quoting long passages...cut them short....

ciao (good method @varun.....thnx)