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

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).

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?

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.

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

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.

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/p^k]. 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).

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).

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).

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.

Thanks a lot!!

Thanks sir. I missed that.

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 .. ??