@DespicableA said:find the remider 21^12-12^21
21^12-12^21 mod 12
21^12 mod 12
7^12 *3^11 mod 4
1*-1=-1
hence 3
3*3=9
9??
@DespicableA said:find the remider 21^12-12^21 when divided by 12
we just need to find 21^12 mod 12, which is 9 always using cyclicity
9
PLZ HELP...
Aakash, bhanu and sam work on a project for some days ,after they complete 1/4th of the work sam takes a break,For 7 days only akash and bhanu work on a project.After that sam relieves both of them.He completes the project in 5 days.bhanu works 50% faster then aakash
and sam alone can complete the project in 20 days,How long would bhanu take to finish the entire project????ans:23 1/3
@Subhashdec2 said:21^12-12^21 mod 1221^12 mod 127^12 *3^11 mod 4 1*-1=-1hence 33*3=99??
i have a doubt..y do u multiply again with 3??
four horses are tethered at four corners of a square plot of side 14 metres so that the adjacent horses can just reach one another .there is a small circular pond of area 20m^2 at the centre..find the ungrazed area??
@pavimai said:four horses are tethered at four corners of a square plot of side 14 metres so that the adjacent horses can just reach one another .there is a small circular pond of area 20m^2 at the centre..find the ungrazed area??
196-49pi-20
196-174=22?
@pavimai said:four horses are tethered at four corners of a square plot of side 14 metres so that the adjacent horses can just reach one another .there is a small circular pond of area 20m^2 at the centre..find the ungrazed area??
14*14-(22/7*7*7)-20
22
1> RIghtmost non zero digit of 15!
2>last two digits of 2^2^2003
@ani6
Let the wrk be of 1 unit..Thus, Rate of Sam= 1/20 unit/days..
Let Aakash alone complete the same wrk in t days..Thus, his rate of wrk = 1/t units/days..
Thus, Rate of Wrking of Bhanu= 3/2*1/t units/day..
Nw, Amt of wrk completed in 7 days= 7*1/t*(1 + 3/2) = 7/t*(5/2) = 35t/2 units..
Wrk already completed= 1/4 + 35t/2 units..Thus, Wrk left for Sam = 3/4 - 35t/2 units..
Given, [3/4 - 35t/2]/1/20 = 5..., or, t = 35 days..
Thus, Bhanu wrking alone will take 2/3rd the time taken by Aaskash alone to complete the wrk..
Thus, Time taken by Bhanu= 2/3*35 = 23 1/3 days..
@amresh_maverick said:1> RIghtmost non zero digit of 15!2>last two digits of 2^2^2003
15!
let n=5a+b
15=5(3)+0
a=3
b=0
right most non zero digit=last digit of[2^a*R(a!)*R(b!)]
===last digit of [2^3*6*1]
=8
how many natural numbers are there with 24 factors where 2,3,5 are the only prime factors??
how many natural numbers are there with 24 factors where 2,3,5 are the only prime factors??
@pavimai -300 is a 1st no having 24 factors where 2,3,5, are the only prime factors,is there any range in btw we want to find out natural nos like 1 to 1000???
@amresh_maverick
Here 15! = 2^11*3^6*5^3*7^2*11*13
Removing 2^3*5^3..we r left with 2^8*3^6*7^2*11*13..Thus, Unit Digit= 6*9*9*1*3 = 8...
For Second it is 02?
@saket.soni05 said:@pavimai -300 is a 1st no having 24 factors where 2,3,5, are the only prime factors,is there any range in btw we want to find out natural nos like 1 to 1000???
range is not mentioned..
Options
36
27
12
9
@pavimai said:how many natural numbers are there with 24 factors where 2,3,5 are the only prime factors??
9?
N = 2^a*3^b*5^c
and (a+1)(b+1)(c+1) = 24
So 2 * 3 * 4 -----6 numbers
or 2* 2 * 6 ------3 numbers
regards
scrabbler
N = 2^a*3^b*5^c
and (a+1)(b+1)(c+1) = 24
So 2 * 3 * 4 -----6 numbers
or 2* 2 * 6 ------3 numbers
regards
scrabbler
@pyashraj said:@amresh_maverickHere 15! = 2^11*3^6*5^3*7^2*11*13Removing 2^3*5^3..we r left with 2^8*3^6*7^2*11*13..Thus, Unit Digit= 6*9*9*1*3 = 8...For Second it is 02?
first correct
2nd OA : 56
2nd OA : 56
@pavimai said:how many natural numbers are there with 24 factors where 2,3,5 are the only prime factors??
Shld be 9..
Let N=2^p*3^q*5^r..Nw, (p+1)(q+1)(r+1)= 24..Since the presence of each prime factors are mandated in the question..We have to c in hw many ways 24 can be written as a product of 3 factor with each prime factor been greater than 1..
Thus, (p+1)(q+1)(r+1)= 2*2*6..Here a total of 3 distinct value of N can be achieved..
Again, (p+1)(q+1)(r+1)=2*4*3..Here a total of 6 distinct value of N can be achieved..
Hence, a total of 9 values...