1wow tht was so awesum prophet sir..........ill giv u a pink hehe
12 Answers
341can u tell me the source of the question. because:
1.it is wrong
2. this about the 10th time i have seen the problem in this form
3. It is more correct to say that \frac{72!}{(36!)^2} - 1 is divisible by 73 (that is a miscellaneous problem in Bernard & Child)
3i was very confusd on seeing the pbm.i wastd more than half of a hr,but i cant come to the conclusion dat it is rong.
341We have 72! = 36! (73-36)(73-35)...(73-1) = 36! (73k+36!) = 73m+(36!)^2
i.e. 72! - (36!)^2 is divisible by 73.
\Rightarrow (36!)^2 \left(\frac{72!}{(36!)^2} - 1 \right) is divisible by 73.
Now, we can prove that \frac{72!}{(36!)^2} is an integer.
We use the fact that the product of n consecutive integers is divisble by n!.
72! = (1 X 2 X 3 X...X 36) X (37 X 38 X 39 X...X72)
Both the bracketed expressions are hence divisible by 36!, so that 72! is divisible by (36!)2
Further, (36!)2 is not divisible by 73 (obvious)
So, if \Rightarrow (36!)^2 \left(\frac{72!}{(36!)^2} - 1 \right) is divisible by 73 and (36!)2 is not divisible by 73, we must have \frac{72!}{(36!)^2} - 1 divisible by 73
341If you know number theory, then the path is a bit easier.
You invoke Wilson's Theorem ( If p is a prime (p-1)!+1 is divisible by p), to say that 73! \equiv -1 \bmod 73
Further, a corollary to this theorem (it follows from one approach of the proof) if p is of the form 4k+1, then \left[\left(\frac{p-1}{2}\right)!\right]^2 \equiv -1 \bmod p
So, we can write, 72! \equiv (36!)^2 \bmod 73
Since 36! is prime to 73, we can divide by (36!)2 and obtain
\frac{72!}{(36!)^2} \equiv 1 \bmod 73
Notice, here we werent concerned with proving that \frac{72!}{(36!)^2} is an integer.
39well , sir is that the only way possible to do it?
anyways good work.
3hey bargave do u know any other method. if so can u do it also.
i proved the first method sir.
Thanq for posting the soln.
i donno number theory sir.is it necessary for jee.if so i will study dat .pls do reply.
341i guess so. both the posts are equivalent. If you observe carefully the number theory result that i quoted proceeds in the same way as worked out in the first post.
3i dont know nething abt number theory sir.If u know ne link dat starts from the basics of number theory can u give it to me.
341you do not need much, if at all. knowing up to Fermat's theorem is sufficient.
Elementary Number Theory by Burton is a good start. Try downloading it.