[warrior]

Originally Posted by vinit456

@ Warrior, Kindly confirm this...
Using divisibilty of 11

(1+3+5+7+9)=25
(2+4+6+ 8_ )=20

now taking the 2 digits nos side by side

Sum of digits in the even places is
45*10=450
odd places..
45*9=405

=(450+20)-(405+25)
=470-430
=40
Possible mistake
I took even-odd

Actually its odd - even, so lets say
=430-470
=-40
so -40%11 is -7
Hence 4???

VA

so dude when will you take even - odd or when will u take odd-even

[vinit456]

Originally Posted by warrior

so dude when will you take even - odd or when will u take odd-even

It's actually odd-even.... Always

I intially made a mistake

Although it's obvious that if the difference turns of to be 0 it doesnt matter

But apne voh wala question mein it's needed

[nagi_nov3]

Originally Posted by Varun Khullar

actually i wrote..
17^-1mod 25 =3..
okay the conceptof inverses goes as this..
consider
a^p/n .. where aand p are coprime.. well do they need to be..(think on that...)
now we used euler number concept to destroy the power..
or we use fermat little theorem.. then we can use remainder theorem and all.. right..
or use chinese remainder theorem..
instead of that.. we can use inverse..
think .. of 2001 ..
last three digits are 001 ..
so its gives remainder as 1.. when divide by 1000
so... its gives the inverse of 3..
667 ..667 =23*29.. so we can't can any inverse of.. this number..
this concept can used to find the last digts easily..
1)31^78/123..find the remainder

now the inverse of 31 is .. 4.. for 123..
31*4 =123+1 .. so..
31^-2 can be written as 4^2 .. and gives 16..

consider other questions.. think in terms of inverses.. its requires tables.. and thinking in terms of factors always..
like from the chocies u will be able to figure out the inverses.. easily..
its pretty similiar to CRM.. crm is a process.. inverses are intuitive..

thanks a lot buddy,

so for
31^78/123
EU for 123 = 80
means 31^-2 = 4^2 = 16

so if a. i mod p =1 then we can take a=1/i any where

[Varun Khullar]

Originally Posted by nagi_nov3

thanks a lot buddy,

so for
31^78/123
EU for 123 = 80
means 31^-2 = 4^2 = 16

so if a. i mod p =1 then we can take a=1/i any where

yah..

[marijuana_user]

Originally Posted by reachmonil

Okies, heres a intersting problem to clear up basics. It was given to me by IdiotR.

Thee number - 50! (Read 50 factorial)

1. Whats the last non-zero digit?
2. The number of 'zeroes'?
3. If 50! is written in base 8 system, how many zeroes would it have?

Enjoy!

JUNTA...!!!!!
50!..will have
12 zeroes
last non zero digit as 2....
doubt abt d last part...

siMILAR QUES WAS AKSED IN pt MOCK CAT ON 3RD sEP....
DECIMAL VALUE OF 100! IS WRITTEN IN ANOTHER BASE v..D NO. OF ZEROES IN D CONVERTED NO IS SAME AS D ORIGINAL NO.

DEN WAT IS D MAX VALUE V CN TK..???
720 30 120 16

WAT IS D MIN VALUE v CN TK...???
2 5 9 16

pT MOCK CAT HAD VERY TOUGH QUANT FRM MY PARANOMA....
dI LAST TWO QUES WERE D RGHT QUES 2 BE ATTEMPTED...SO 20 MARKS 4M DRE
N ENGLISH WAS ALWAYS N IS VERY TOUGH 4 ME...WEN I READ CHOICE a IT SEEMS 2 BE D ANSWER..WEN I READ CHOICE b IT SEEMS 2 BE D ANSWER..WAT DO I DO..????

[anupam_khanna]

[QUOTOriginally Posted by reachmonil
Okies, heres a intersting problem to clear up basics. It was given to me by IdiotR.

Thee number - 50! (Read 50 factorial)

1. Whats the last non-zero digit?
2. The number of 'zeroes'?
3. If 50! is written in base 8 system, how many zeroes would it have?

Enjoy![/QUOTE]

well the no of 0's is simple... it comes out to be as 12

if it is written in base 8 the no of zeros will still be 12.... (if this is correct then i'll explain how...)

still trying to calculate the last non zero digit of 50!

[nagi_nov3]

Originally Posted by anupam_khanna

Thee number - 50! (Read 50 factorial)

1. Whats the last non-zero digit?

Enjoy!

here u go

1.2.3.4.5.6.7.8.9.10
1.2.3.4.15.6.7.8.9.20
1.2.3.4.25.6.7.8.9.30
1.2.3.4.35.6.7.8.9.40
1.2.3.4.45.6.7.8.9.50


so apart from 5's and 10's we can take any number mod 10 as last digit
12 mod 10 = 2 (last digit)

with 5 we need to be careful
5 *2 = 10, and 5*12 = 60

1.2.3.4.5.6.7.8.9.10
1.2.3.4.5*3.6.7.8.9.10*2
1.2.3.4.5*5.6.7.8.9.10*3
1.2.3.4.5*7.6.7.8.9.10*4
1.2.3.4.5*9.6.7.8.9.10*5

So we left with
10!.1.1 = 10!. 1.(2/2)
10!. 3.2 = 10!. 3. (4/2)
10!. 5.3 = 10!. 5. (6/2)
10!.7.4 = 10! . 7. (8/2)
10!.9.5 = 10!. 9. (10/2)

meams

we have

(10!)^6 / 2^5

we know for 10! = 8

so 8^6 /2^5 = 2^13 = 2. 4^12 = 2.6^6 = 2

now try for 100!

[anupam_khanna]

Originally Posted by nagi_nov3

here u go

1.2.3.4.5.6.7.8.9.10

now try for 100!
...........

naagi sahab aap to bhagwaan ho...:grab:
sirji 1 baar vo inverse wala funda acche se explain kar do please

[Varun Khullar]

Originally Posted by anupam_khanna

naagi sahab aap to bhagwaan ho...:grab:
sirji 1 baar vo inverse wala funda acche se explain kar do please

well .. 100!
can written as 8^10 20!/2^10 .. so we get 4 as answer..
summit ji ka method dekho..
z(100) =4^10 z(20)
z(20) =4^2 z(4)
4^10 = 6..
4! is 24 .. and 6*4 = 4.. so answer is 4.. summit bhai ka method bhi sahi(i mean mast hai) hai..

inverse ka funda.. a*i mod p =1 then i is the inverse of a when divided by p......
are yaar question main lagane ka.. na

[nagi_nov3]

Originally Posted by anupam_khanna

nagi sahab aap to bhagwaan ho...:grab:
sirji 1 baar vo inverse wala funda acche se explain kar do please

we shud say thanks to Varun for this inverse funda

the thing is

apart from applying Eulers funda we shud try to apply this also

suppose we need to find last 3 digits of 3^398

3^398 mod 1000

EN for 1000 = 400

means we need to find 3^-2 mod 1000

try to find a number x such that 3x mod 1000 =1

means try for (1000k+1)/3
for k=2 = 2001/3 = 667

so inverse of 3 is 667

so u can replace 3 = 1/667

so our problem reduced to
667^2 mod 1000

otherwise we need to go by long way