MY CART (5)

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R

There are no items in this cart.
Continue Shopping

X

“Hurray! You have won the gift voucher. Redeem your points now."

X

                   


the highest prime less than 50, that divides the binomial coefficient 100C50 ? a) 37, b) 31, c)47 d)43


3 years ago

Share

Answers : (13)

                                        

100! contains 47 and 94, 50!*50! contains 47 twice. Therefore, answer is 47

3 years ago
                                        

Hi Rituparna,


 


For the greatest prime number, say p.


3p < 100


 


Because 100C50 = 100!/(50!*50!)


Each of the 50! would cancel the prime numbers in 100! twice. So there has to be one more prime for it to divide 100C50.


So 3p<100. Only possibility p=31.


 


Option B.


 


Regards,


Ashwin (IIT Madras).

3 years ago
                                        

yeah, the anwer is 31.... ur logic is right .... actually there is one more method using Legendre theorem..... would u know it by any chance?

3 years ago
                                        

I am guessing 33 should be the required answer. Because exponent of 33 in 100! is 3 and in 50!^2 is 2. But that isnt in option.

3 years ago
                                        

yeah the answer is b), thanx, but there is one more way using Legendre .... would u know it by any chance?

3 years ago
                                        

I have never heard of Legendres theorem. But is it this one by any chance (i dont know the name)


 


Exponent of p in n! = [n/p] + [n/p2] + [n/p3] + ....


 


If yes, then i have used this.

3 years ago
                                        

Hi Rituparna,


 


Yes you can use Legendre Theorem here.


It states the highest power of a prime number "p" in N! would be = Σ[N/pk]. where k = 1,2,..... and [.] denotes greatest integer function.


For example the power of 5 in 100! would be [100/5] + [100/25] + [100/125] +.... = 20+4 = 24. ie 100! will have 5^24.


 


So in this case check the prime number, for which you get the exponent to be three or more.


Only 31 will satisy that condition.


 


Hence option (B) using Legendre Theorem !!.


 


Regards,


Ashwin (IIT Madras).

3 years ago
                                        

Hi Jit Mitra,


 


Yes that is one of Legendre Theorem.


But it is applicable only on prime numbers.


And 33 = 3x11 is not a prime number.


 


And hence 31 is the largest prime number.


 


Regards,


Ashwin (IIT Madras).

3 years ago
                                        

If one needs to get the exponent of 33, then one has to split it into prime numbers first and then get the exponent.


 


33=3x11.


 


exp of 3 would be [100/3] + [100/9] + [100/27] + [100/81] = 33+11+3+1 = 48.


And exp of 11 would be [100/11] = 9.


 


So exp of 33 would be the lesser of the two.


Hence exponent of 33 in 100! would be 9.


 


Regards,


Ashwin (IIT Madras).

3 years ago
                                        

ys the ans wuld b 47.


thats bcoz, 100C50 = (51*52*53*......*100)/50!   and hence any num having a factor in series in numerator wuld divide.


also as per given terms in ques, 47 wuld be the correct 1.

3 years ago
                                        

Thanks a lot!!

3 years ago
                                        

Thanks sir. I missed that.

3 years ago
                                        

actually i m a bit weak at these binomial cofficients nd all .


so i have a kind request to ashwin sir ..... sir can u pls give some 4-5 relevant nd good questions based on the same or a bit diffrent concept .. ??

3 years ago

Post Your Answer

More Questions On Algebra

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!!
Click Here for details
if a,b,c,d are positive real numbers such that a/3=a+b/4=a+b+c/5=a+b+c+d/6 then a/b+2c+3d
 
 
Let a/3 = a + b / 4 = a + b + c / 5 = a + b + c + d/ 6 = k then, a = 3k, b = k, c = k, d = k a/b + 2c + 3d = 3k/k + 2k + 3k = 3k/6k = 1/2
 
Y RAJYALAKSHMI 11 months ago
 
hey post raledu kada
 
kasilaxmi 11 months ago
 
hello chandu.....
 
farzana 11 months ago
i cant get an idea to solve problems in complex numbers . suggest me an idea
 
 
Hi could you please give more details about the kind of doubts you have? The best method of course would be practicing with solved examples first and then slowly moving onto unsolved ones...
 
Kalyani Jayachandran Menon 12 days ago
 
i cannot do the sums on ` this equation represents -----------------`
 
Sarathchandra Lavu 11 days ago
question is in image
 
 
Step 1: C3 = C3 – C2*x, Step 2: R3 = R3 + R2, Step 3: C1 = C1 – 50*C2, Step 4: Expand the determinant You will get Dx = 2x(-50) + 500, Sum(Dx) = -500
 
Akshay 11 days ago
What is the Value of sin A if BC=5cm And AB=10cm, given that triangle is right angled at B?
 
 
sin A = opp side / hypotenuse = bc /ac= 5/ root125
 
Rahul Jiji George 26 days ago
the point with position vectors 60i +3j, 40i- 8j and i-52 j are collinear if a =-40 a=40 a=20 none of these
 
 
answer is 40 for collinear we can write AB=X (BC) WHERE a,b,c ARE THREE GIVEN VECTORES AND ON COMPARING THE COEFFICIENTS OF i and j we get the value of a to be 40 approve if useful
 
ng29 3 months ago
 
hey mansi check the question again as there is no a in all the given vectors and post the que again to get it solved
 
ng29 3 months ago
 
the point with position vectors 60i+3j, 40i-8j and ai-52j are collinear if a =-40 a=40 a=20 none of these
 
mansi dabriwal 3 months ago
View all Questions »