shanks4mba Saysoption 2? 18081? just 9*2009:)
I guess so.. I got the same answer. It should be the one.
But is it so simple? What's that, some kind of a thumb rule or sumthin?
what would happen if the power was raised to some higher number?
I guess so.. I got the same answer. It should be the one.
But is it so simple? What's that, some kind of a thumb rule or sumthin?
what would happen if the power was raised to some higher number?
I guess so.. I got the same answer. It should be the one.
But is it so simple? What's that, some kind of a thumb rule or sumthin?
what would happen if the power was raised to some higher number?
just chk for smaller no. like 9*9 = 81-->8+1 = 9-->9*1
99*99 = 9801 = 9+8+1 = 18-->9*2
u'll get the flow:)
I guess so.. I got the same answer. It should be the one.
But is it so simple? What's that, some kind of a thumb rule or sumthin?
what would happen if the power was raised to some higher number?
u can create ur thumb rule by jus doing some simple calcu, in this case
9^2 = 81 => sum(81) = 9, which is 1*9
99^2 = 9801 = sum(9801) = 18 is 2*9
999^2 = 997001 = sum(997001) = 27 is 3*9
hence sum(9999.....2009 times ^2) = 2009*9
This ques is from arun sharma q44, LOD III
N=202 X 20002 X 200000002 X 2 00 ...2(fifteen zeroes) X 2000....(31 zereos)2.
find sum of digits of multiplication: :banghead:
a.112 b.160 c.144 d.cant be determined
ans= 160, my take 144
This ques is from arun sharma q44, LOD III
N=202 X 20002 X 200000002 X 2 00 ...2(fifteen zeroes) X 2000....(31 zereos)2.
find sum of digits of multiplication: :banghead:
a.112 b.160 c.144 d.cant be determined
ans= 160, my take 144
here is the app !!
taking 2's common ..
2*2*2*2*2 *101*10001*100000001*10000...... so on !!
i.e 32 * 10101010101010101010........... 32 ones and 31 zeroes..
i.e 32032032032032032..............................
that is 32 comes 32 times ......
3+2 * 32 --> 5 *32 --> 160
AN EASY ONE
X = (10000 10001 10002 10003) + 1 and X = Y2, where Y is a natural number.
What is the value of Y?
OPTIONS 1)100000001 2)102030201 3)102000201 4)103000301 5)None of these
ANS 5
AN EASY ONE
X = (10000 10001 10002 10003) + 1 and X = Y2, where Y is a natural number.
What is the value of Y?
OPTIONS 1)100000001 2)102030201 3)102000201 4)103000301 5)None of these
ANS 5
Getting 100030001 i.e. None of these.........
As y^2 = (10000 * 10001 * 10002 * 10003) + 1 or
y^2 - 1 = 10000 * 10001 * 10002 * 10003 or
(y+1)(y-1) = 10000 * 10001 * 10002 * 10003
Pairing.........
(y+1)(y-1) = (10000 * 10003) * (10001 * 10002) or
(y+1)(y-1) = 100030000 * 100030002 i.e.
y = 100030001..........
AN EASY ONE
X = (10000 10001 10002 10003) + 1 and X = Y2, where Y is a natural number.
What is the value of Y?
OPTIONS 1)100000001 2)102030201 3)102000201 4)103000301 5)None of these
ANS 5
FOR QSTN OF THIS TYPE USE THIS trick:-
firstly we shud know multiplying 4consecutive digits +1=perfect square
n*(n+1)*(n+2)*(n+3)=^2
so here n=10000
ans is 100030001. option 5 ans
Another new creation from pg's den
Please have a look if not already 
http://www.pagalguy.com/prep/
What will be the remainder when
999 ^9 is divided by 99^9 ?
This ques is from arun sharma q44, LOD III
N=202 X 20002 X 200000002 X 2 00 ...2(fifteen zeroes) X 2000....(31 zereos)2.
find sum of digits of multiplication: :banghead:
a.112 b.160 c.144 d.cant be determined
ans= 160, my take 144
4040404
808080808080808
16161616161616161616161616161616
N= 32323232323232323232323232323232
32323232323232323232323232323232
sum of digits (N)=5*16*2
hence ans is 160
what will be the remainder when
999 ^9 is divided by 99^9 ?
Wrong post
What will be the remainder when
999 ^9 is divided by 99^9 ?
=(999/99)^9
=9^9
999^9 mod 99^9
=> 9^9 * 111^9 mod 9^9 * 11^9
=> 111^9 mod 11^9
=> (110+1)^9 mod 11^9
=> + + .... + mod 11^9
Taking Term By Term
=> mod 11^9 = 0
=> mod 11^9 = 9c1 * 11^8
=> mod 11^9 = 9c2 * 11^7
....
....
=> mod 11^9 = 1
So remainder => 9^9 *
{I've divided 9^9 before so multiplying Again to the remainder I've got}
Comments/Suggestions Welcome
999^9 mod 99^9
=> 9^9 * 111^9 mod 9^9 * 11^9
=> 111^9 mod 11^9
=> (110+1)^9 mod 11^9
=> + + .... + mod 11^9
Taking Term By Term
=> mod 11^9 = 0
=> mod 11^9 = 9c1 * 11^8
=> mod 11^9 = 9c2 * 11^7
....
....
=> mod 11^9 = 1
So remainder => 9^9 *
{I've divided 9^9 before so multiplying Again to the remainder I've got}
Comments/Suggestions Welcome
nice method but a bit complicated..
can u plz help by any simple method ..
thanx...
What will be the remainder when
999 ^9 is divided by 99^9 ?
no offences 2 d brainies who hav solved above...but wit a diif answer..
my take is a bit diff..
(90)^9..
wts d OA???do post it..
What will be the remainder when
999 ^9 is divided by 99^9 ?
999^9 = 111^9 * 9^9
999^9 = 11^9 * 9^9
so we get 111^9/11^9 ...
remainder is 1
999 ^9 is divided by 99^9 ?
999^9 = 111^9 * 9^9
999^9 = 11^9 * 9^9
so we get 111^9/11^9 ...
remainder is 1
note: 111^2 / 11^2 = 100,
naga pls check divisth post for the question, so that we can discuss further
pls tell me about eulers theorem used n number system???hw can i calculate d eulers value???is dr a table fr it???pls explain r else provide me a link where i cn study eulers theorem
pusparghya Sayspls tell me about eulers theorem used n number system???hw can i calculate d eulers value???is dr a table fr it???pls explain r else provide me a link where i cn study eulers theorem
Before Posting your Query Buddy, Please Search The Forums
Euler's Theorem - http://www.pagalguy.com/forum/cat-and-related-discussion/43764-2009-ncr-dream-team-26.html#post1685280
Chinese Remainder Theorem - http://www.pagalguy.com/forum/cat-and-related-discussion/43743-the-shout-boxers-team-09-a-24.html#post1680098
Hab Phun...
Hi Guys,
Could you please explain how to approach following two types of problems?
1. Find last two digits of 65*29*37*63*71*87
2. What is the greatest power of 12 in 123!
Thanks,
Rohit.