Official Quant thread for CAT 2013

ravi6389 · February 25, 2013, 4:43pm

for this problemAll possible 6 digit numbers, in each of which the digits occur in non - increasing order (from left to right) are written in increasing order. What is the 500th number in this sequence ?OPTIONS1) 6321002) 6221103) 6331104) 642100

Yes OA is 3

logrhythm · February 25, 2013, 4:43pm

@rkshtsurana said:
Q. (21) ^ 21^21^21^21......(21 times) remainder by 25 ?P.S - After a long time posting here, Feel so good here..so a question from my side

e(25) = 20

21^21^... %20 = 1

21^1%25 = 21...

scarecrow28 · February 25, 2013, 4:44pm

@rkshtsurana said:
Q. (21) ^ 21^21^21^21......(21 times) remainder by 25 ?P.S - After a long time posting here, Feel so good here..so a question from my side

21?

scrabbler · February 25, 2013, 4:44pm

@rkshtsurana said:
Q. (21) ^ 21^21^21^21......(21 times) remainder by 25 ?P.S - After a long time posting here, Feel so good here..so a question from my side

21?

Hi after a long time :)

regards
scrabbler

logrhythm · February 25, 2013, 4:44pm

@ravi6389 said:
for this problemAll possible 6 digit numbers, in each of which the digits occur in non - increasing order (from left to right) are written in increasing order. What is the 500th number in this sequence ?OPTIONS1) 6321002) 6221103) 6331104) 642100Yes OA is 3

check a few pages back.... @scrabbler and I had tried solving it....thoda mushkil ques hai...

ravi6389 · February 25, 2013, 4:46pm

@Logrhythm yes i just checked scrabble's solution..was really very nice.

viewpt · February 25, 2013, 4:46pm

@rkshtsurana said:
Q. (21) ^ 21^21^21^21......(21 times) remainder by 25 ?P.S - After a long time posting here, Feel so good here..so a question from my side

21 eulers

deadly · February 25, 2013, 4:47pm

@rkshtsurana said:
Q. (21) ^ 21^21^21^21......(21 times) remainder by 25 ?P.S - After a long time posting here, Feel so good here..so a question from my side

21??

saurav205 · February 25, 2013, 4:47pm

@rkshtsurana said:
Q. (21) ^ 21^21^21^21......(21 times) remainder by 25 ?P.S - After a long time posting here, Feel so good here..so a question from my side

21 aaygea.

E(25) = 20

21^21^21.....%20 = 1

hence 21 remainder.

scrabbler · February 25, 2013, 4:47pm

@Logrhythm said:
e(25) = 2021^21^... %20 = 1 21^1%25 = 21...

@rkshtsurana

Mera method thoda different tha...orally kiya...

Looked at the pattern of last 2 digits for 21 ke powers:
21, 41, 61, 81, 01....
so 21^21^21.... = 21^5k+1 ends in 21 itself. Now this number divided by 100 will leave remainder 21 so with 25 too it will leave 21...

regards
scrabbler

viewpt · February 25, 2013, 4:48pm

@rkshtsurana said:
Q. (21) ^ 21^21^21^21......(21 times) remainder by 25 ?P.S - After a long time posting here, Feel so good here..so a question from my side

kaahey bhai..itni khushi??

deadly · February 25, 2013, 4:49pm

@scrabbler said:
@rkshtsuranaMera method thoda different tha...orally kiya...Looked at the pattern of last 2 digits for 21 ke powers:21, 41, 61, 81, 01....so 21^21^21.... = 21^5k+1 ends in 21 itself. Now this number divided by 100 will leave remainder 21 so with 25 too it will leave 21...regardsscrabbler

ye hamare oral quant guru hain

viewpt · February 25, 2013, 4:49pm

Q:R= x= 73^79 div 100 find R?

logrhythm · February 25, 2013, 4:49pm

@scrabbler said:
@rkshtsuranaMera method thoda different tha...orally kiya...Looked at the pattern of last 2 digits for 21 ke powers:21, 41, 61, 81, 01....so 21^21^21.... = 21^5k+1 ends in 21 itself. Now this number divided by 100 will leave remainder 21 so with 25 too it will leave 21...regardsscrabbler

i fit euler's at every possible jagah where i can...but aapke method hamesha thode better hi hote hai...sahi hai.. :)

logrhythm · February 25, 2013, 4:52pm

@viewpt said:
Q:R= x= 73^79 div 100 find R?

73^80%100 = 1

73^79*73%100 = 1

73*x%100 = 1

so x = 37...

ilovetorres · February 25, 2013, 4:52pm

@scrabbler said:
@rkshtsuranaMera method thoda different tha...orally kiya...Looked at the pattern of last 2 digits for 21 ke powers:21, 41, 61, 81, 01....so 21^21^21.... = 21^5k+1 ends in 21 itself. Now this number divided by 100 will leave remainder 21 so with 25 too it will leave 21...regardsscrabbler

bhai idhar 81 ke baad hi 61 aata hai na.. 😛 kabhi kabhi oral examm mmein bhi fail ho sakte hai :P

saurabhlumarrai · February 25, 2013, 4:54pm

@Logrhythm said:
i fit euler's at every possible jagah where i can...but aapke method hamesha thode better hi hote hai...sahi hai..

thanks to you...aapke hi euler har jagah lagane ki wajah se hum bhi euler me master ho gaye hain...thanks again bro :)

mailtoankit · February 25, 2013, 4:57pm

@viewpt said:
TPQ: A writes all the nos. frm 1 to 1000 on paper in order. Find the 2883rd digit written by him?

7 ?

1 - 9 = 9 digits

10 - 99 = 180 digits

100 - 999 = 2700 digits

9 + 180 + 2700 - 6 = 2883..which comes out to be = 997

rkshtsurana · February 25, 2013, 4:57pm

@viewpt said:
Q:R= x= 73^79 div 100 find R?

E(100) = 100*4/5* 1/2 = 40
so

73^40 = 1 mod 100

73^80 = 1 mod 100

or 73 N -1 is divisible by 100

ie 73N -1 = 100p

N = 37 remainder

deadly · February 25, 2013, 4:57pm

@viewpt said:
Q:R= x= 73^79 div 100 find R?

27??