@anupam001 said:QA to fodte ho boss....VA kam hai mera 55 tha ye waala
yaar VA ka koi upaye batao... Kharab toh hai hi upar se half the time concentration bhi nahi rehta karte time !! :(
@ash1615 said:yaar VA ka koi upaye batao... Kharab toh hai hi upar se half the time concentration bhi nahi rehta karte time !!
mera khud hi gol hai...kuch pata nhi rehta...mai aajkal 1-2 attempt kam kar ke har question pe 15-20 second jyada spend karta hun...thodi accuracy improve hui hai
I've also started with TIME sectionals:
QADI advanced #3: 70 (24-2)
VALR advanced #2: 53(20-7)
@DEBJITNAG said:I've also started with TIME sectionals:QADI advanced #3: 70 (24-2)VALR advanced #2: 53(20-7)
arre sir...aap bht dino baad aaye iss thread me :)...good going :)
Late on the home ground! Apologies!
Congratulations to the new team! :)
@kpraveen010 said: which of these sections are easy to get score ? paragraph questions scramble questions vocabulary last sentence is deleted sentence correct plz give reply .....
wats scramble question??
in order it shld be
vocabulary
PJ
PC
SC
yaar ek cheez batao...if in a probability questn its given that either of them win ...sumthin lyk that...toh in this do v consider the case when both of dem win ...
@anishnambisan said: Late on the home ground! Apologies!Congratulations to the new team!
kaisa chal rha hai fms me sir :)
@nits2811 said: yaar ek cheez batao...if in a probability questn its given that either of them win ...sumthin lyk that...toh in this do v consider the case when both of dem win ...
should take....koi question hai to bata 1 baar
exactly aisa question toh nhi hain..par dimag mein doubt aagya...i read sumwhere that v dont...
@nits2811 said: exactly aisa question toh nhi hain..par dimag mein doubt aagya...i read sumwhere that v dont...
hmmm....baakiyon ko batane dete hain fir
how to find the last digit/last 2 digits in case we have the number of form a^b! eg (13)^103!
@@abhi_14
I might not be the intended audience for your question, but felt like answering, so kindly bear with me.. :)
Look, we solve last digit/last 2 digit qsns based on cyclicity of nos...
Last digit qsns could be easily solved by considering the cyclicity of the no N (Cosider qsn of the format N^(abc)! .
A bigger factorial will always be a multiple of 4 and a cyclicity of 4 will always give you result for the last digit. (As abc!= abc*(abc-1)*.....*5*4*3*2*1 ).
So the power will be of the form 4k. (ie abc!=4k or 103!=4k in our case)
Ultimately the qsn will boil down to last digit of N^4k..
Which in our case is 13^4k..
Since 13^4 gives unit digit 1*(3*3*3*3=9*9=81= 1 unit digit), so 13^4k will also give unit digit 1.
Extend the same logic to other qsns..
Now coming to the last 2 digits, remember that we solve last 2 digits of N^x by using last 2 digits of N and getting some pattern which repeats due to power on N..
e.g. last 2 digits of 211^321 = last 2 digits of 11^321... which could be easily solved by using binomial theorem..
Coming to your qsn of last 2 digits of 13^103!..
Remember that last 2 digits of 13^4=61
Now 103! will have 25+5+1 = 31 no of 0's..
So 103! = ...................000000000 31 times..
Even if we write 103! as (4 * Something ) that something will contain 29 zeros as 4*25 =100 which will add to the existing no of zeroes in 103!.
So ultimately our qsn becomes last 2 digits of 13^(4*xyzvw...000000...000 29 zeroes), which is nothing else but 61^(xyzvw...000000...000 29 zeroes) which will give 01 as last 2 digits as 61^ (xyzvw...000000...000 29 zeroes) = (60+1)^(xyzvw...000000...000 29 zeroes) = higher powers of 60 + 60*(xyzvw...000000...000 29 zeroes) + 1*1. [ Expanded through binomial theorem]
Use the same logic for other qsns..
Hope that helps.. :)
I might not be the intended audience for your question, but felt like answering, so kindly bear with me.. :)
Look, we solve last digit/last 2 digit qsns based on cyclicity of nos...
Last digit qsns could be easily solved by considering the cyclicity of the no N (Cosider qsn of the format N^(abc)! .
A bigger factorial will always be a multiple of 4 and a cyclicity of 4 will always give you result for the last digit. (As abc!= abc*(abc-1)*.....*5*4*3*2*1 ).
So the power will be of the form 4k. (ie abc!=4k or 103!=4k in our case)
Ultimately the qsn will boil down to last digit of N^4k..
Which in our case is 13^4k..
Since 13^4 gives unit digit 1*(3*3*3*3=9*9=81= 1 unit digit), so 13^4k will also give unit digit 1.
Extend the same logic to other qsns..
Now coming to the last 2 digits, remember that we solve last 2 digits of N^x by using last 2 digits of N and getting some pattern which repeats due to power on N..
e.g. last 2 digits of 211^321 = last 2 digits of 11^321... which could be easily solved by using binomial theorem..
Coming to your qsn of last 2 digits of 13^103!..
Remember that last 2 digits of 13^4=61
Now 103! will have 25+5+1 = 31 no of 0's..
So 103! = ...................000000000 31 times..
Even if we write 103! as (4 * Something ) that something will contain 29 zeros as 4*25 =100 which will add to the existing no of zeroes in 103!.
So ultimately our qsn becomes last 2 digits of 13^(4*xyzvw...000000...000 29 zeroes), which is nothing else but 61^(xyzvw...000000...000 29 zeroes) which will give 01 as last 2 digits as 61^ (xyzvw...000000...000 29 zeroes) = (60+1)^(xyzvw...000000...000 29 zeroes) = higher powers of 60 + 60*(xyzvw...000000...000 29 zeroes) + 1*1. [ Expanded through binomial theorem]
Use the same logic for other qsns..
Hope that helps.. :)
@utsav7986 Thanks bro for the lucid xplanation....gt the concept :)
PS: while posting this query i didn had any intended audience in my mind :P, it was posted on the forum so anybody can rply...
Thanks again!!!
ATB.....:)
@utsav7986 said: @@abhi_14I might not be the intended audience for your question, but felt like answering, so kindly bear with me.. Look, we solve last digit/last 2 digit qsns based on cyclicity of nos... Last digit qsns could be easily solved by considering the cyclicity of the no N (Cosider qsn of the format N^(abc)! . A bigger factorial will always be a multiple of 4 and a cyclicity of 4 will always give you result for the last digit. (As abc!= abc*(abc-1)*.....*5*4*3*2*1 ). So the power will be of the form 4k. (ie abc!=4k or 103!=4k in our case) Ultimately the qsn will boil down to last digit of N^4k.. Which in our case is 13^4k.. Since 13^4 gives unit digit 1*(3*3*3*3=9*9=81= 1 unit digit), so 13^4k will also give unit digit 1. Extend the same logic to other qsns..Now coming to the last 2 digits, remember that we solve last 2 digits of N^x by using last 2 digits of N and getting some pattern which repeats due to power on N..e.g. last 2 digits of 211^321 = last 2 digits of 11^321... which could be easily solved by using binomial theorem..Coming to your qsn of last 2 digits of 13^103!..Remember that last 2 digits of 13^4=61Now 103! will have 25+5+1 = 31 no of 0's.. So 103! = ...................000000000 31 times.. Even if we write 103! as (4 * Something ) that something will contain 29 zeros as 4*25 =100 which will add to the existing no of zeroes in 103!.So ultimately our qsn becomes last 2 digits of 13^(4*xyzvw...000000...000 29 zeroes), which is nothing else but 61^(xyzvw...000000...000 29 zeroes) which will give 01 as last 2 digits as 61^ (xyzvw...000000...000 29 zeroes) = (60+1)^(xyzvw...000000...000 29 zeroes) = higher powers of 60 + 60*(xyzvw...000000...000 29 zeroes) + 1*1. [ Expanded through binomial theorem]Use the same logic for other qsns.. Hope that helps..
but suppose v use euler's in this case and divide the number by 100 and find the remainder..den the last two digits wld be 13 only..
euler's quotient wld b 40...now 40 divides 103! completely...so it shld be 13 only..
wats wrng here??
udt confy starting in a few minutes.. please be available guys on gmail
@raven_007 said: udt confy starting in a few minutes.. please be available guys on gmail
Mujhey bhi add kar dena usmein. ID is same as username. :)