How many factor of N are 4m+1 type?
N = 5^3*2^4*3^5
@sparklingaubade said:N = 5^3*2^4*3^5How many factor of N are 4m+1 type?
No power of 2 should be selected, Any power of 5, and only even power of 3, then the factor will have a remainder of 1 when divided by 4
=> number of such factors = (4)*(1)*(3) = 12 ?
@sparklingaubade said:N = 5^3*2^4*3^5How many factor of N are 4m+1 type?
12?
No power of 2 should come
3 mod 4 = -1
5 mod 4 = 1
Hence all powers of 5 can come and even powers of 3 can come
Hence , 4* 3 =12 ways ..
@YouMadFellow I have got 11 factors
5,25,125,3^2 ,3^4 ,3^2*5 , 3^2*5^2 , 3^2*5^3
3^4*5 , 3^4*5^2 , 3^4*5^3
Which factor i m missing here.?????
5,25,125,3^2 ,3^4 ,3^2*5 , 3^2*5^2 , 3^2*5^3
3^4*5 , 3^4*5^2 , 3^4*5^3
Which factor i m missing here.?????
@Aman.Malhotra said:@YouMadFellow I have got 11 factors 5,25,125,3^2 ,3^4 ,3^2*5 , 3^2*5^2 , 3^2*5^33^4*5 , 3^4*5^2 , 3^4*5^3Which factor i m missing here.?????
1
@jain4444 said:There are 3 candles with their lengths in the ratio 1 : 2 : 3 (Every other dimension is the same for all the candles). They are lit in such a way that when the second candle has been lit, the first candle had been reduced to half its original length & when the third candle is lit, the second candle is half its original length. The total time taken for all the candles to totally burn out is 9 hours. Assume that the candles are lit in increasing order lengths. In how much time does the longest candle completely burn?
6?
A bag contains 1 Red, 2 Blue, 3 Green and 4 Yellow balls. Three balls are drawn from the bag randomly, one at a time, and the ball replaced into the bag every time. What is the probability that the three balls drawn are of different colours?
@hiteshkhurana82 said:A bag contains 1 Red, 2 Blue, 3 Green and 4 Yellow balls. Three balls are drawn from the bag randomly, one at a time, and the ball replaced into the bag every time. What is the probability that the three balls drawn are of different colours?
(1*2*3 + 1*3*4 + 1*2*4 + 2*3*4)/1000 = 50/1000 = 1/20
EDIT:- oops forgot to multiply by 3!
@hiteshkhurana82 said:A bag contains 1 Red, 2 Blue, 3 Green and 4 Yellow balls. Three balls are drawn from the bag randomly, one at a time, and the ball replaced into the bag every time. What is the probability that the three balls drawn are of different colours?
P(red) = 1/10
P(blue) = 2/10
P(green) = 3/10
P(yellow) = 4/10
P(three balls of different colors) = RBG + RBY + RGY + BGY
= 3!*(1/10^3)[(1*2*3) + (1*2*4) + (1*3*4) + (2*3*4)]
= 6*[ 6 + 8 + 12 +24 ]/1000 = 3/10 ?
@hiteshkhurana82 said:A bag contains 1 Red, 2 Blue, 3 Green and 4 Yellow balls. Three balls are drawn from the bag randomly, one at a time, and the ball replaced into the bag every time. What is the probability that the three balls drawn are of different colours?
3/10
Total no of ways in which the balls can be removed from the bag = 10*10*10 = 1000
Total combinations possible = RGB,RGY,RBY,GBY
RGB = (1*3*2/1000)*3!
RGY = (1*3*4/1000)*3!
RBY = (1*2*4/1000)*3!
GBY = (2*3*4/1000)*3!
Total = (50/1000)*3! = 3/10
Please correct me if i am wrong
YouMadFellow soumitrabengeri
I witnessed THE GOD making a mistake for the first time!
N = 7^7*19^4*13^5
How many factor of N are 6m+1 type?
How many factor of N are 6m+1 type?
@hiteshkhurana82 said:A bag contains 1 Red, 2 Blue, 3 Green and 4 Yellow balls. Three balls are drawn from the bag randomly, one at a time, and the ball replaced into the bag every time. What is the probability that the three balls drawn are of different colours?
RBG = 1C1*2C1*3C1 = 6
RGY = 1C1*3C1*4C1 = 12
BGY = 2C1*3C1*4C1 = 24
RBY = 1C1*2C1*4C1 = 8
favorable cases = 50*3!
total cases = 10^3
so , prob. = 50*6/10^3
@sparklingaubade said:N = 7^7*19^4*13^5How many factor of N are 6m+1 type?
Same principle as used earlier
7 mod 6 = 1
19 mod 6 =1
13 mod 6 = 1
-> Every factor is in the form 6m + 1 => 8*5*6 = 240 ?
@hiteshkhurana82 said:A bag contains 1 Red, 2 Blue, 3 Green and 4 Yellow balls. Three balls are drawn from the bag randomly, one at a time, and the ball replaced into the bag every time. What is the probability that the three balls drawn are of different colours?
(1*2*3+2*3*4+3*4*1+1*2*4)/1000=3/10
@hiteshkhurana82 said:A bag contains 1 Red, 2 Blue, 3 Green and 4 Yellow balls. Three balls are drawn from the bag randomly, one at a time, and the ball replaced into the bag every time. What is the probability that the three balls drawn are of different colours?
RGB=1/10*2/10*3/10
BGY=2/10*3/10*4/10
GYR=3/10*4/10*1/10
YRB=4/10*1/10*2/10
Multiply by 3! for different arrangements of the 3 combinations and add we get
3/10
BGY=2/10*3/10*4/10
GYR=3/10*4/10*1/10
YRB=4/10*1/10*2/10
Multiply by 3! for different arrangements of the 3 combinations and add we get
3/10