Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R

There are no items in this cart.
Continue Shopping
Get instant 20% OFF on Online Material.
coupon code: MOB20 | View Course list

Get extra R 440 off
USE CODE: CART20

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

```

4 years ago

Share

### Answers : (13)

```										100! contains 47 and 94, 50!*50! contains 47 twice. Therefore, answer is 47
```
4 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,
```
4 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?
```
4 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.
```
4 years ago
```										yeah the answer is b), thanx, but there is one more way using Legendre .... would u know it by any chance?
```
4 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.
```
4 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,
```
4 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,
```
4 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,
```
4 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.
```
4 years ago
```										Thanks a lot!!
```
4 years ago
```										Thanks sir. I missed that.
```
4 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 .. ??
```
4 years ago

# Other Related Questions on Algebra

If alpha is a real root of the equation ax 2 +bx+c and beta is a real root of equation -ax 2 +bx+c. Show that there exists a root gama of the equation (a/2)x 2 +bx+c which lies between...

 Ajay 4 months ago

Small Mistake in last para posting again..............................................................................................................

 Ajay 4 months ago

We have Similarly, So if P(x) = a/2 x 2 +bx +c, then and are off opposite sign and hence there must exist a root between the two numbers.

 mycroft holmes 4 months ago
In the listed image can you tell me how beta*gamma = 2 ….. . . .. ??

The value of gamma is still not correct, either printing mistake or you gave me wrong value. The correct value of gamma is below

 Ajay 3 months ago

Thankyou so much............................. …......................................................................!

 Anshuman Mohanty 3 months ago

Yes sorry..... . . . .it is not so clear.. ok the values are beta = α + α^2 + α^4 and gamma = α^3 + α^5 + α^7

 Anshuman Mohanty 3 months ago
prove that root 3 is irrational number?

P and Q are natural numbers, and hence in the prime factorization of P 2 and Q 2 , every prime factor appears to an even power. Now in the equation P 2 =3Q 2 we have a contradiction as LHS...

 mycroft holmes 3 months ago

prove that for all p,q that equation not satisfies. and send post the answer . need proof for that, how do I conclude that is not going to satisfy by directly telling so just post it

 BHOOPELLY SAIKUMAR 3 months ago

WE NEED TO PROVE THAT FOR ALL P,Q SUCH THAT P 2 =3Q 2 DOESN’T EXIST WHERE P,Q BELONGS TO Z. so justify ur answer with proper reason ….

 BHOOPELLY SAIKUMAR 3 months ago
sin^2 6°-sin^2 12°+sin^2 18°-sin^2 24°......15 solve it Urgent

Ajay, the complete qution isSolution is sin^2 6°-sin^2 12°+sin^2 18°-sin^2 24°..... upto 15 terms. sin 78°=0 sin 42°+sin 54°+ sin 66°+ + sin 18° sin 6°+ where )=0.5 (your required answer),...

 Kumar 2 months ago

Not any people get my answer why. You can no give answer my question I am join this site

 Vivek kumar 4 months ago

Hello If you want to get the solution quick you should post your question in clear manner. Its not clear what you wnat us to solve, and what does 15 at the end of question means?

 Ajay 4 months ago
Solve: (sin theta+cosec theta)^2 + (cos theta +sec theta)^2- (tan^2 theta + cot ^2 theta)^2

What needs to be solved here ? The question is incomplete....................................................................

 Ajay 4 months ago

i don’t know how to do this...............................................................................................

 Saravanan 24 days ago

this is the question :: Solve: (sin theta+cosec theta)^2 + (cos theta +sec theta)^2 - (tan^2 theta + cot ^2 theta)

 Naveen Shankar 4 months ago
proove that {cot^a(secA-1)}/1+sinA = sec^2A pls help me solve this problem and thak you

I hope you want to prove follwing identity. If yes then this not correct question as this can be never be true . For example put A = 45, LHS and RHS are not equal.

 Ajay 5 months ago

Hi Gaurang the question still does not seem to be correct. Can you post the screenshot of question attachment

 Ajay 5 months ago

Yes ajay sir I didnt type it correct so it was amistke please help in soving this one and thankyou very much

 Gaurang Kotasthane 5 months ago
View all Questions »

• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: R 15,000
• View Details
Get extra R 3,000 off
USE CODE: CART20

Get extra R 440 off
USE CODE: CART20

More Questions On Algebra