Divide all no by 3...1,4,8,11,22..
Now, 1 + 3 = 4, 4*2=8, 8+3=11,11*2=22
Next is 22+3=25..and 25*3=75
@sunnychopra89 said:A number has exactly 15 composite factors. What can be the minimum and maximum number of prime factors of this number?
15 + 2 = 17 (min 1 prime)
(a+1)(b+1) - 2 - 1 = 15
(a+1)(b+1) = 18 (max 2 prime)
So min 1, max 2
(a+1)(b+1) - 2 - 1 = 15
(a+1)(b+1) = 18 (max 2 prime)
So min 1, max 2
@sunnychopra89 said:A number has exactly 15 composite factors. What can be the minimum and maximum number of prime factors of this number?
Minimum is no doubt 1.
Maximum will be 2.
Maximum will be 2.
A cargo truck missed a goods train. The train leaves 25 seconds before the truck reaches the station. The train travels at 10 m/s while the maximum speed at which the truck can travel is 12 m/s. The next station is 10 km away. Approximately, what is the longest period for which the truck has to wait for the train at the next station?
15 composite factors.. Plus 1 will be factor itself.(1- neither prime nor composite)
Now, total 16 factors.
case-1 for min consider number to be 2^15. so only one prime factor = 2.
case-2 for max. consider no to be 2*3*5*7 ..so 4 prime factors = 2,3,5,7
@Amrofa said:A cargo truck missed a goods train. The train leaves 25 seconds before the truck reaches the station. The train travels at 10 m/s while the maximum speed at which the truck can travel is 12 m/s. The next station is 10 km away. Approximately, what is the longest period for which the truck has to wait for the train at the next station?
141.67s?
In 25 sec, the train has travelled 250m.
Distance to be covered by train = 9.75km
9750/10 - 10000/12 = 975 - 833.33 = 141.6667 s
In 25 sec, the train has travelled 250m.
Distance to be covered by train = 9.75km
9750/10 - 10000/12 = 975 - 833.33 = 141.6667 s
@anuragu79 said:@sunnychopra8915 composite factors.. Plus 1 will be factor itself.(1- neither prime nor composite)Now, total 16 factors.case-1 for min consider number to be 2^15. so only one prime factor = 2.case-2 for max. consider no to be 2*3*5*7 ..so 4 prime factors = 2,3,5,7
total will be composite factors + 1 + prime factors :)
@grkkrg said:15 + 2 = 17 (min 1 prime)
(a+1)(b+1) - 2 - 1 = 15
(a+1)(b+1) = 18 (max 2 prime)
So min 1, max 2
Dude Can u plz explain this... I'm finding it difficult to understand.
Let a, b and c are positive integers. Let i2 = €“1.
If c = (a + bi)3 €“ 107i, then find c
If c = (a + bi)3 €“ 107i, then find c
@sunnychopra89
A number has total factors in three parts
Total factors = Prime fac + Composite fac + 1
We added 1 because '1' is factor of every number,which is neither prime nor composite.
Now total number of factors except the prime one this number will have 15+1=16.
Now we will see the options.
lets try '2'.
totAL factors are 18.
18 can be written as (8+1)*(1+1) or (5+1)*(2+1).So we can have thAt number possible.
let's try '3'
total factors 19.
We have to write 19 as product of three numbers which is not possible as it's prime.
let's try '4'
total factors= 20
We have to write 20 as a product of four numbers greater than one which is not possible.
same case with 21 As it can't be written as product of 5 numbers.
Hence maximum prime factors this number can have is 2.
A number has total factors in three parts
Total factors = Prime fac + Composite fac + 1
We added 1 because '1' is factor of every number,which is neither prime nor composite.
Now total number of factors except the prime one this number will have 15+1=16.
Now we will see the options.
lets try '2'.
totAL factors are 18.
18 can be written as (8+1)*(1+1) or (5+1)*(2+1).So we can have thAt number possible.
let's try '3'
total factors 19.
We have to write 19 as product of three numbers which is not possible as it's prime.
let's try '4'
total factors= 20
We have to write 20 as a product of four numbers greater than one which is not possible.
same case with 21 As it can't be written as product of 5 numbers.
Hence maximum prime factors this number can have is 2.
@sunnychopra89 said:Dude Can u plz explain this... I'm finding it difficult to understand.
15 composite factors + 1 + prime factors
Let the number be N = p^a * q^b * ...
where p,q,... are prime
Then the total number of factors = (a + 1) (b + 1) .....
Now if we have one prime factor
N = p^16
Number of factors = 16 + 1 = 17 (which includes 1 and p)
Number of composite factors = 17 - 2 = 15
If we have two prime factors
N = p^2 * q^5
Number of factors = 3 * 6 = 18 (which includes 1,p,q)
Number of composite factors = 18 - 3 = 15
We cannot have more than 2 prime factors. (checked for 3,4,5 prime factors)
Hence min 1, max 2.
Let the number be N = p^a * q^b * ...
where p,q,... are prime
Then the total number of factors = (a + 1) (b + 1) .....
Now if we have one prime factor
N = p^16
Number of factors = 16 + 1 = 17 (which includes 1 and p)
Number of composite factors = 17 - 2 = 15
If we have two prime factors
N = p^2 * q^5
Number of factors = 3 * 6 = 18 (which includes 1,p,q)
Number of composite factors = 18 - 3 = 15
We cannot have more than 2 prime factors. (checked for 3,4,5 prime factors)
Hence min 1, max 2.
@SBRPhoenix said:Let a, b and c are positive integers. Let i2 = €“1.If c = (a + bi)3 €“ 107i, then find c
c=a^3 -ib^3 + 3a^2bi -3ab^2 -107i
c=(a^3 - 3ab^2) + i(3a^2b-b^3-107)
equating the imaginary part to 0
3a^2b-b^3=107
b(3a^2b-b^2)=1*107
if b=1 then a=6
if b=107 then a no longer an integer.
thus, b=1 and a=6
and c=198
EDIT: made a silly earlier
c=(a^3 - 3ab^2) + i(3a^2b-b^3-107)
equating the imaginary part to 0
3a^2b-b^3=107
b(3a^2b-b^2)=1*107
if b=1 then a=6
if b=107 then a no longer an integer.
thus, b=1 and a=6
and c=198
EDIT: made a silly earlier
@SBRPhoenix said:Let a, b and c are positive integers. Let i2 = €“1.If c = (a + bi)3 €“ 107i, then find c
a = 6
b = 1
c = 108
c = a^3 - b^3i + 3abi ( a + bi) - 107i
c = a^3 - b^3i + 3a^2bi - 3ab^2 - 107i
c = (a^3 - 3ab^2) + i(3a^2b - b^3 - 107)
3a^2b - b^3 - 107 = 0
3a^2b - b^3 = 107
b(3a^2 - b^2) = 1 * 107
b = 1
a = 6
c = 216 - 108 = 108
b = 1
c = 108
c = a^3 - b^3i + 3abi ( a + bi) - 107i
c = a^3 - b^3i + 3a^2bi - 3ab^2 - 107i
c = (a^3 - 3ab^2) + i(3a^2b - b^3 - 107)
3a^2b - b^3 - 107 = 0
3a^2b - b^3 = 107
b(3a^2 - b^2) = 1 * 107
b = 1
a = 6
c = 216 - 108 = 108