The pg Kolkata Dream Team

[soumyadut]

Originally Posted by rik_12

please...please....drop that h in "shinchan"...
evn anir had d same prob last year...ppl called him "anirbhan" and his DT tee had "Kolkatha"..
so drop it fast

arey yaar drop both the "h"-s

aaj se uska naya naam

SIN-CAN

paap ka bhandar

[techguyz2008]

The remainder whn 2^2+22^2+222^2+.......(222.......49twos)^2 is divided by 9 is :

(a)2
(b)5
(c)6
(d)7


Question is frm ArunSharma Page:51 Qno. 43.

Right answer is 6.... i hav solved this problm...... but need to knw frm u guys... how u guys approach this kindaa questions..... reply asap...thnx in advance.........aftr seeing couple frm u all .. i will post mine........


@SINCHAN......... sry bro...... only 1 'h' frm 2day onwards......

[yudhajeet]

Originally Posted by techguyz2008

The remainder whn 2^2+22^2+222^2+.......(222.......49twos)^2 is divided by 9 is :

(a)2
(b)5
(c)6
(d)7


Question is frm ArunSharma Page:51 Qno. 43.

Right answer is 6.... i hav solved this problm...... but need to knw frm u guys... how u guys approach this kindaa questions..... reply asap...thnx in advance.........aftr seeing couple frm u all .. i will post mine........


@SINCHAN......... sry bro...... only 1 'h' frm 2day onwards......

Am gettin 4 as the answer


2^2+22^2+222^2+.......(222.......49twos)^2 = 2^2+4^2+6^2+8^2+ 1^2+4^2+6^2+8^2 (11 times) +1^2

Now, 1^2+4^2+6^2+8^2 divided by 9 gives 0 remainder.

2^2+4^2+6^2+8^2+ the last 1^2 divided by 9 gives remainder 4

So, remainder 4. Dunno where am goin wrong.

Correct me puys!!

[shriyan]

Originally Posted by yudhajeet

Am gettin 4 as the answer


2^2+22^2+222^2+.......(222.......49twos)^2 = 2^2+4^2+6^2+8^2+ 1^2+4^2+6^2+8^2 (11 times) +1^2

Now, 1^2+4^2+6^2+8^2 divided by 9 gives 0 remainder.

2^2+4^2+6^2+8^2+ the last 1^2 divided by 9 gives remainder 4

So, remainder 4. Dunno where am goin wrong.

Correct me puys!!

Here's How i solved it...
rem[{2^2 + 22^2 + 222^2 +......+ (222...49 times)^2},9]
=rem[ 2^2{1^2 + 11^2 + 111^2 + .... +(111...49 times)^2} , 9 ]


Now...any no. whn divided by 9 leaves a remainder whch is equal to the remainder whn the sum of digits of that particular no. is divided by 9.

Hence...rem(1,9) = 1 rem(1^2,9)= 1^2
rem(11,9) = 2 rem(11^2,9) = 2^2
rem(111,9) = 3 rem(111^2,9) = 3^2
rem(1111,9) = 4 rem(1111^2,9) =4^2

... .....
rem(11111111,9)=8 rem(11111111^2,9) = 8^2
rem(111111111,9)=0 rem(111111111^2) = 0

rem(1111111111,9)=1 rem(11...10times^2,9)=1^2..

Cyclic....Order(cycle of 9)...


So in 49, 9 goes 5 times...

Hence REM = 2^2[{1^2 + 2^2 + .......... +8^2}x5 + 1^2(for 46) + 2^2(for 47) + 3^2(for 4 + 4^2(for 49) ]

=REM[4{5( 8 x 9 x 17)/6 + 30} , 9]

=REM [ 4 x 5 x (12 x 17 +6) , 9]

= REM [ 20 x (12 x 17 +6) , 9]

= REM [ 2 x (3 x -1 +6) , 9]

=REM [2x3 , 9]

=6....ANS..

[ronildutta]

Originally Posted by yudhajeet

Am gettin 4 as the answer


2^2+22^2+222^2+.......(222.......49twos)^2 = 2^2+4^2+6^2+8^2+ 1^2+4^2+6^2+8^2 (11 times) +1^2

Now, 1^2+4^2+6^2+8^2 divided by 9 gives 0 remainder.

2^2+4^2+6^2+8^2+ the last 1^2 divided by 9 gives remainder 4

So, remainder 4. Dunno where am goin wrong.

Correct me puys!!

mine is a tedious process...............
takin cue from yudhajeet.......................

here remainder comes as.......
2^2+22^2+222^2+.......(222.......49twos)^2 = 2^2+4^2+6^2+8^2+ 1^2+3^2+5^2+7^2 (5 times) + 2^2+4^2+6^2+8^2 =

5*(17*9*8 )/6 + 2^2(4*5*9)/6........(sum of 1st n square terms....ie n(n+1)(2n+1)/6.......)
= 5*17*3*4 + 4*2*5*3 ............here wen we divide by 9 both the terms.................we get 3+3.........ie remainder=6................

hope it helps.............

[shriyan]

Originally Posted by yudhajeet

Am gettin 4 as the answer


2^2+22^2+222^2+.......(222.......49twos)^2 = 2^2+4^2+6^2+8^2+ 1^2+4^2+6^2+8^2 (11 times) +1^2

Now, 1^2+4^2+6^2+8^2 divided by 9 gives 0 remainder.

2^2+4^2+6^2+8^2+ the last 1^2 divided by 9 gives remainder 4

So, remainder 4. Dunno where am goin wrong.

Correct me puys!!

Ur approach is abs Correct...but u're makin a mistake 1.e.

2222^2/9 leaves a rem = 8^2
22222^2/9 leaves a rem = 1^2 = (-eight)^2...
222222^2/9 leaves a rem = 3^2 = (-6)^2
2...7tms^2 / 9...rem= 5^2 = (-4)^2..
2...8times^2 / 9...rem = 7^2 = (-2)^2..
2...9tms^2 / 9...rem = 0

*U're nt cnsdrn trms div by 9...

So...REM[ (2^2 + 4 ^ 2 + 6^2 + 8^2) x 11 , 9 ]
=REM [ { 4 + (-2) + 0 + 1 } x 2 } , 9]
=REM [ 6,9] = 6....Ans

.....

[shriyan]

Yudha...was jus skimmin thrugh A. Sharma...
Got Stuck...
128^1000 divided by 153..
Wht's the Remainder??

[shriyan]

I never Got my AIM 13 Solutions...
Can't Solve DI 81-85...
Can TechGuyz or Captain Charisma...could u plzz explain??

[techguyz2008]

Sry fr posting befre the analysis session of AIM911 cuz by the time i will join the analysis session, math section analysis will prbly be over......


Q43. of AIM911..... TIME authrity has assigned this prblm as a difficult 1.... but this prblm shud nt take more thn 30 secs to solve.... prbly my approach is much effective thn AIMCAT solutions ...

Total there is 7 equations as far the questions.... but in my approach u just need 2 equations to solve this ...

A4 + A5 + A6 = -1;
A5 + A6 + A7 = 1;
from this 2 we get :
A4 - A7= -2

There is only 1 option tht satisfies A4-A7=-2

So the required value is (-5/3,1/3)

[techguyz2008]

Originally Posted by shriyan

I never Got my AIM 13 Solutions...
Can't Solve DI 81-85...
Can TechGuyz or Captain Charisma...could u plzz explain??

Here are the solutions of AIM913 Q81-85.....

Q81.Min possible no. f persons to attend the festival :

Here our approach shud be to minimize the no. f persons watchng multiple movies.....

TZP(Really a must watch movie): Only this movie = 100
Multiple movie as well as TZP = 110+90+50-100=150
CDI(Best acting evr by KING KHAN): Only this movie = 40
Multiple movie as well as CDI = 110+60+30-40=160
GURU(AISH was looking terrific): Only this movie = 30
Multiple movie as well as GURU = 60+50+40-30=120
THE NAMESAKE(Love the bedscene by TABU): Only this movie = 20
Multiple movie as well as TN = 30+70+70-20=150

No. of persons who watched single movie = 100+40+30+20=190
Min. no. of persons who watched multiple movies= max.(150,160,120,150)=160

So ans is = 190+160=350


Q82.Here we need to maximize no. f persons aged over 60 watching only a single movie ...

TZP: 50<100 so single movie=50
CDI: 30<40 so single movie=30
GURU: 40>30 so single movie=30 and multiple movie 10
TN: 30>20 so single movie=20 and multiple movie 10

So we can see 10 persons aged more thn 60 saw GURU and TN bth

So total no. f persons = 50+30+30+20+10=140

Q83.Here we need to maximize multiple movie watchers for less thn 60 age group..
NAME LHS RHS Multiple
movies
TZP : ((110+90)-(100-50))=150 RHS=50(only TZP)
CDI : ((60+110)-(40-30))=160 RHS=10(only CDI)
GURU : ((50+60)-(30<40))=110 RHS= 0
TN : ((70+70)-(20<30))=140 RHS=0

Ans = 50+10+max.(150,160,110,140)
=220



Q84.As we need to find max. no f persons watching all 4 movies so ans will b min.(150,160,120,150)=120

Q85.Here our approach is tht all multiple movie watchers hav saw xactly 2 movies
[(50+110+90-100)+(30+60+110-40)+(40+50+60-30)+(30+70+70-20)]/2=580/2=290



Tried to make it as simple as possible... hope u like it...


CheerzZ