@Torque024 said:Yep, approach matters. Post it.
I didn't think prime numbers could ever follow any sort of pattern :/
Also the last digit will fluctuate in cycles of 3,7,5,1. Any number ending with 5 other than 5 itself is not prime.
@Gaul said:Find the greatest term in the expansion of หลก3(1 + (1/ หลก3))^20
20C5*1/3^4 , 5th terms ?
@bullseyes said:Find area of square if 2x -3y + 7 = 0 and 6x - 9y + 8 = 0 represent 2 sides of square
7/rt13-8/3rt13=rt13/3
area of square = 13/9
A contractor undertakes to dig a canal 10Km long in 180 days and employs 40 men. After 60 days he finds that only 2.5 km of the canal has been completed. To complete the work on the scheduled time how many men does he have to increase ?
10
20
25
15
explnation pls..
10
20
25
15
explnation pls..
@ankita14 said:I didn't think prime numbers could ever follow any sort of pattern :/Also the last digit will fluctuate in cycles of 3,7,5,1. Any number ending with 5 other than 5 itself is not prime.
when you take first one as 29,,then the others end up in 9 only..it gets difficult in this case..at what point it gets non prime :\
@ankita14 said:0
@sujamait @Estallar12 @mailtoankit
If An belongs to integerA0 = p, where p > 0 is a prime number,An+1 = 2An + 1, for n = 0, 1, 2, .....How many values of p exist such that sequence consists entirely of prime numbers
Adding 1 to the recurrence relation, we get An+1 + 1 = 2(An + 1).
Hence An + 1 = (2^n)*(A0 + 1).
Substituting A0 = p, we obtain an = (2^n)*p + (2^n โ 1).
If p = 2, then A1 = 5 is prime, and so, without loss of generality, we may assume that p is odd.
Consider An congruent to 2n โ 1 (modulo p).
Then, since 2 and p are relatively prime, by Fermat's Little Theorem, 2^(pโ1) congruent to 1 (mod p).
Hence Apโ1 congruent to 0 (mod p).
Since Apโ1 > p, it follows that Apโ1 is composite.
Therefore, there is no value of p such that the above sequence consists entirely of prime numbers.
If An belongs to integerA0 = p, where p > 0 is a prime number,An+1 = 2An + 1, for n = 0, 1, 2, .....How many values of p exist such that sequence consists entirely of prime numbers
Adding 1 to the recurrence relation, we get An+1 + 1 = 2(An + 1).
Hence An + 1 = (2^n)*(A0 + 1).
Substituting A0 = p, we obtain an = (2^n)*p + (2^n โ 1).
If p = 2, then A1 = 5 is prime, and so, without loss of generality, we may assume that p is odd.
Consider An congruent to 2n โ 1 (modulo p).
Then, since 2 and p are relatively prime, by Fermat's Little Theorem, 2^(pโ1) congruent to 1 (mod p).
Hence Apโ1 congruent to 0 (mod p).
Since Apโ1 > p, it follows that Apโ1 is composite.
Therefore, there is no value of p such that the above sequence consists entirely of prime numbers.
@catter2011 said:A contractor undertakes to dig a canal 10Km long in 180 days and employs 40 men. After 60 days he finds that only 2.5 km of the canal has been completed. To complete the work on the scheduled time how many men does he have to increase ?10202515explnation pls..
20?
@Estallar12 said:Only ONE when p = 2. ?
How come? 2, 5, 11, 23, 47, 95....not prime...or am I reading something wrong in the question?
(In fact if 2 were prime 5 and 11 and all the others would also satisfy...)
I would go with zero for the simple reason that if there is such a p, then every number in the series starting with p would also satisfy the condition and hence the answer would be infinite.
regards
scrabbler
(In fact if 2 were prime 5 and 11 and all the others would also satisfy...)
I would go with zero for the simple reason that if there is such a p, then every number in the series starting with p would also satisfy the condition and hence the answer would be infinite.
regards
scrabbler
@catter2011 said:A contractor undertakes to dig a canal 10Km long in 180 days and employs 40 men. After 60 days he finds that only 2.5 km of the canal has been completed. To complete the work on the scheduled time how many men does he have to increase ?10202515explnation pls..
.25W=40*60
w=40*60*4
remaining w=.75w
=4*6*4*75=remaining days*mens
=4*6*4*75=120*M; M=60
Men to increase=60-40=20
w=40*60*4
remaining w=.75w
=4*6*4*75=remaining days*mens
=4*6*4*75=120*M; M=60
Men to increase=60-40=20
@catter2011 said:A contractor undertakes to dig a canal 10Km long in 180 days and employs 40 men. After 60 days he finds that only 2.5 km of the canal has been completed. To complete the work on the scheduled time how many men does he have to increase ?10202515explnation pls..
.25W=40*60
w=40*60*4
remaining w=.75w
=4*6*4*75=remaining days*mens
=4*6*4*75=120*M; M=60
Men to increase=60-40=20
w=40*60*4
remaining w=.75w
=4*6*4*75=remaining days*mens
=4*6*4*75=120*M; M=60
Men to increase=60-40=20
@catter2011 said:A contractor undertakes to dig a canal 10Km long in 180 days and employs 40 men. After 60 days he finds that only 2.5 km of the canal has been completed. To complete the work on the scheduled time how many men does he have to increase ?10202515explnation pls..
Already done: 2.5 in 60. To do: 7.5 in 120. So 3 times the work in 2 times the time means 3/2 times the men => 60 men => 20 extra.
regards
scrabbler
@catter2011 said:A contractor undertakes to dig a canal 10Km long in 180 days and employs 40 men. After 60 days he finds that only 2.5 km of the canal has been completed. To complete the work on the scheduled time how many men does he have to increase ?10202515explnation pls..
60d*40m = .25 w
120d*(40+k)m = .75 w = .75 * 2400/.25
4800 + 120k = 7200
k= 20
@catter2011 said:@mailtoankit sahi hein.. explanation.
work1/M1*D1=work2/M2*D2
2.5 KM of canal completed by 40 men in 60 days...so rest 7.5km of canal will be completed by (40+x)men in 120days (days left 180-60=120)
2.5/60*40=7.5/(40+x)*120....x=20
@scrabbler said:How come? 2, 5, 11, 23, 47, 95....not prime...or am I reading something wrong in the question?(In fact if 2 were prime 5 and 11 and all the others would also satisfy...)I would go with zero for the simple reason that if there is such a p, then every number in the series starting with p would also satisfy the condition and hence the answer would be infinite.regardsscrabbler
Good logic:clap:, but there was a more than 8 option too in all fairness :p
@catter2011 said:A contractor undertakes to dig a canal 10Km long in 180 days and employs 40 men. After 60 days he finds that only 2.5 km of the canal has been completed. To complete the work on the scheduled time how many men does he have to increase ?10202515explnation pls..
20??
40 men --->1/4th work---> 60 days
Each man does 1/9600th of the work in 1 day
60 men will complete 3/4th work in 120 days-->meets the deadline
Hence,20 more men will be needed
@Brooklyn said:how did u calculate distance b/w d lines??
Two parallel lines can always be represented in the form,
ax + by + c1 =0 and
ax + by + c2=0
Here, multiply the first eqn by 3, u get
6x - 9y + 21=0 and
6x - 9y + 8=0
Distance between two parallel lines= |c1 - c2|/ sqrt(a^2 + b^2)
This gives the side of the square. Square it to get the area as 13/9
ax + by + c1 =0 and
ax + by + c2=0
Here, multiply the first eqn by 3, u get
6x - 9y + 21=0 and
6x - 9y + 8=0
Distance between two parallel lines= |c1 - c2|/ sqrt(a^2 + b^2)
This gives the side of the square. Square it to get the area as 13/9
@catter2011 20 ?
40 men work for 60 days and complete 2.5 Km canal , thus 2400 Man-days is required for 1/4th of work. So, for remaining 3/4th work 3*2400 man-days will be required.
3*2400=120*X
X=60 men
So, additional 20 men will be required.
@ankita14 said:Good logic, but there was a more than 8 option too in all fairness
True hence it was a calculated risk ๐ Would take the chance in the exam...what's life without a little risk?
regards
scrabbler
regards
scrabbler
Find the sum of co-primes less thn 14