@anantn said:@shattereddream3/8
its 3/4 not 3/8 you have done a mistake
@ayushverma said:What is the probablity of two personsamong the four having birthday in he same month?A. 75%B. 50%C. 33%
b. 50 %
@ayushverma said:What is the probablity of two personsamong the four having birthday in he same month?A. 75%B. 50%C. 33%
(12*12*12*4C2)/(12^4) = 50%.
@ayushverma said:What is the probablity of two personsamong the four having birthday in he same month?A. 75%B. 50%C. 33%
50%?
@ao4mba said:(12*12*12*4C2)/(12^4) = 50%.
Can u explain in a little bit more detail?
I did it like this:
Choose 2 people in 4C2 ways.
Assign a month to them in 12C1 ways.
Now the remaining 2 cannot get that month and they must have distinct months.(Since exactly 2 people have same months as birthday)
So assign months to them in 11*10 ways.
So ans =4C2*12*11*10/12^4
I did it like this:
Choose 2 people in 4C2 ways.
Assign a month to them in 12C1 ways.
Now the remaining 2 cannot get that month and they must have distinct months.(Since exactly 2 people have same months as birthday)
So assign months to them in 11*10 ways.
So ans =4C2*12*11*10/12^4
@ayushverma said:What is the probablity of two persons
among the four having birthday in he same month?
A. 75%
B. 50%
C. 33%
4C2 selecting 2 out of 4 = 6
probablity of two having birthday on same day = 1 * 1/12
therefore, 6 * 1/12 = 1/2
50 %
A number has exactly 32 factors out of which 4 are not composite. Product of these 4 factors(which are not composite) is 30. How many such numbers are possible?
Approach Plz
@pankaj1988 said:A number has exactly 32 factors out of which 4 are not composite. Product of these 4 factors(which are not composite) is 30. How many such numbers are possible?Approach Plz
we can split into 3 primes(2,3,5) and 1 since product is 30
N= 2^a*3^b*5^c
(a+1)(b+1)(c+1)= 32 = 2*4*4 or 2*2*8
so 6 such numbers are possible
@pankaj1988 said:A number has exactly 32 factors out of which 4 are not composite. Product of these 4 factors(which are not composite) is 30. How many such numbers are possible?Approach Plz
6 no.s?
2^a*3^b*5^c
(a+1)(b+1)(c+1) = 32
8*2*2---> 3 numbers
4*4*2---> 3 numbers
total 6 numbers
@pankaj1988 said:A number has exactly 32 factors out of which 4 are not composite. Product of these 4 factors(which are not composite) is 30. How many such numbers are possible?Approach Plz
the prime factors must be 2,3,5 since the product is 30.
32 = 4x4x2 => 3 numbers
32 = 2x2x8 => 3 numbers
total 6 number possible
@pankaj1988 said:A number has exactly 32 factors out of which 4 are not composite. Product of these 4 factors(which are not composite) is 30. How many such numbers are possible?Approach Plz
There are ,as the question mentions, 4 non-composite factors.
One of these factors is 1, rest must be the prime factors of that number.
Product is 30.
Therefore nos are 1,2,3,5
Now there are 3 prime factors of the number and the number is of the form----
(2^p)*(3^q)*(5^r) where p,q,r>=1
Total factors are 32=(p+1)(q+1)(r+1)
Now, 32 can be represented as product of three integers in two ways (none of the three integers can be 1).
32=2*2*8
32=2*4*4
case 1
two of p,q are 1 and the other is 7
total ways are 3
case 2
two of p,q are 3 and the other is 1
total ways are 3
ans=6
One of these factors is 1, rest must be the prime factors of that number.
Product is 30.
Therefore nos are 1,2,3,5
Now there are 3 prime factors of the number and the number is of the form----
(2^p)*(3^q)*(5^r) where p,q,r>=1
Total factors are 32=(p+1)(q+1)(r+1)
Now, 32 can be represented as product of three integers in two ways (none of the three integers can be 1).
32=2*2*8
32=2*4*4
case 1
two of p,q are 1 and the other is 7
total ways are 3
case 2
two of p,q are 3 and the other is 1
total ways are 3
ans=6
@pankaj1988 said:A number has exactly 32 factors out of which 4 are not composite. Product of these 4 factors(which are not composite) is 30. How many such numbers are possible?Approach Plz
6
Find the remainder when (123123123123..................300digits) is divided by 37.
Please explain along with the method?
How many divisors of 360 are not divisors of 540 and how many divisors of 540 are not divisors of 360?
1000 gives a remainder of 1 with 37
123 can be represented as
123 +123*1000+123*1000*1000 and so on
now remainders of individual terms are
123+123+123......and so on till 100 terms
therefore we can find the remainder with
(123*100)mod 37
=(12*26)mod 37
=16
@ytstacks said:Find the remainder when (123123123123..................300digits) is divided by 37.Please explain along with the method?
See if you divide 123 by 37 you will get 12 as remainder. Now your number has 100 groups of 123 serially arranged and repeating....
so each groupd will give remainder of 12. So basically it becomes 12*100 divided by 37
Rem. 1200/37=16
so each groupd will give remainder of 12. So basically it becomes 12*100 divided by 37
Rem. 1200/37=16