and your signature says "have it figured out." :mg:
@Love_CAT said: @Estallar12and your signature says "have it figured out."
But the meaning is different. "it" refers to Life here :P
@Estallar12 said:But the meaning is different. "it" refers to Life here
what if
Life screws you just when you think you have figured it out.
woh to prank tha :P
@Love_CAT said:what ifLife screws you just when you think you have figured it out.woh to prank tha
Yeah, I know it should be like "figured it out" but aisa daala to get the meaning of "it figured out" - Again a Prank see :mg:
@gautam22 said:atleast tell whether its rite or wrong......den it makes point to post d approach.....noone wud like to waste their time on wrong approach..
Done have the OA. :P
@gautam22 said:see first the 2^47 in the 50! will b cancelled by 16^15 ....left will be 2^13.......after division we will hav 1-25 all odd no's and 1-50 prime no's den formed d pairs and multiplied.....sorry ans is wrong instead of multiplying i have added d pairs formed .....sorry
Yeah. Answer will be something else. Doesn't seem doable to me. :|
@Estallar12
@Estallar12 said:Remainder when 50! is divided by 16^15.P.S. on't have the OA. Approach daalna. Don't want any random answers.
16^15=2^60
now 50! has 48 power of 2..so we r left with1.3.5.7.9.11.13..........upto 49/2^12
You can write numerator as (2k1+1)(2k2+1)..........(2k49+1).....
So i gess rem. will be 1..
50!= 2^47*3^22*5^12*7^8*11^4*13^3*17^2*19^2*23^2*29*31*37*41*43*47
since den= 2^60, we cancel out 2^47 in num and den and find remainder when 50! is divided by 2^13.its very cumbersome.so will i leave such questions.:)
since den= 2^60, we cancel out 2^47 in num and den and find remainder when 50! is divided by 2^13.its very cumbersome.so will i leave such questions.:)
@deekshaaneja said: @Estallar1216^15=2^60now 50! has 48 power of 2..so we r left with1.3.5.7.9.11.13..........upto 49/2^12You can write numerator as (2k1+1)(2k2+1)..........(2k49+1).....So i gess rem. will be 1..
50! has 47 powers of 2.
And also, after cancelling out 2^47 we will have 22 powers of 3, 12 powers of 5 and so on. I guess bahut ganda ho jaega aage.
@deekshaaneja said: @Estallar1216^15=2^60now 50! has 48 power of 2..so we r left with1.3.5.7.9.11.13..........upto 49/2^12You can write numerator as (2k1+1)(2k2+1)..........(2k49+1).....So i gess rem. will be 1..
you never know what the product of those odd numbers might lead to. had it been some odd number ^ k , k greater than 13, remainder would have been 1 and final rem would have been 1*2^47. but @Estallar12 ..tu aise ghatiya qsns poocha toh log pareshaan hojayenge!!
@naarto - Sir, someone had asked it to me. I was not able to figure out anything. I thought I might be missing some link and so posted it here. Anyway, this is not a feasible question to solve then. :D
An easy question:A set has n elements. Set B has m elements.
Find the number of bijective( one-one ) onto functions if
a) n
b)n=m
c)n>m
giving a formula doesnt prove the whole point of posting the question. Elucidate with an example.
stats: #supposedly a probable area for a question to be picked from, every year.
@naarto said: An easy question: A set has n elements. Set B has m elements. Find the number of bijective( one-one ) onto functions if a) nb)n=m c)n>m giving a formula doesnt prove the whole point of posting the question. Elucidate with an example. stats: #supposedly a probable area for a question to be picked from, every year.
only for n=m bijective functions will be obtained if I am not wrong...
no of bijective functions = m! =n! .. its like 4 different balls has to go into 4 different boxes such that each box has only 1 ball..
yar kisi ke pas...wordlist hai audio format me?
kisi k pas SIMCATs pade h this year/last year, need it for section-2
@abhi_14 said: kisi k pas SIMCATs pade h this year/last year, need it for section-2
haan shayad yahi pe kisi ne bheja tha... email id Pm karo.. bhejta hun.. 😃