IIT JEE/MAINS MATH (CHANDRA KANT SIR)

A good question on function.. enjoy!!

Good question on probability. Will be helpful in developing better understanding. Enjoy guys!!🤟

Good question based on recursive sequences. Solve and enjoy learning!! 🤟

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Dm your answers

Find all positive integers n such that
3^n - 2^n = 65...(*).
Writing in mod 3,
-(-1)^n = -1, which forces n to be even, say 2m.
(*) becomes 9^m-4^m=65, ie
(3^m+2^m)(3^m-2^m)= 65.
Cases arise:
1) 3^m+2^m=65 &
3^m -2^m=1.
This has no solution.
2) 3^m+2^m=13 &
3^m-2^m =5, which leads to m=2, giving n=4.
Mathematics is a trial & error, like sciences.

A question inspired by jee mains:
If sqrt[(4+2n)/(1+n)] is a rational number ,then least even natural value of n is?
Ans is 48 🙃

Q)Find the longest string of consecutive positive integers which add to 2013. (Pre RMO).
Take a + (a+1)+...+(a+k-1) =2013, giving k(2a+k-1)=4026.
This equation implies :
k | 4026 & k^2 < 4026,
the latter giving k < 63.5(app).
List the divisors of 4026 in ascending order. 61 is the greatest to fit these requirements.
k(max)=61.
The string is 3, 4, 5, ..., 63.
Bounding an unknown is a standard strategy.

"Any fool can do mathematics. All that it needs is perseverance."
( Alonzo Church, an English logician).
Q) Find all primes in the sequence
1,101,10101,1010101,..., in which the digits 1 & 0 alternate, starting, & ending , with 1.
You can easily identify some composites in the sequence.
Thus the 3rd, 6th, 9th,..., terms are divisible by 3 ; the 11th, 22nd, 33rd,..., terms are divisible by 11.
On the other hand,101 is prime.
However, these observations do not complete your investigation.
Let n be the term having k 1's ( & k-1 0's).
The crucial move is
99n= 99...9 ( 2k times)
= 10^(2k) -1
= ( 10^k -1)(10^k +1)...(*)
If n were prime, then n must divide one of 10^k-1 & 10^k + 1 ( Why ?)
In case of the former, via(*), 10^k +1 divides 99, which is impossible.
In case of the latter, via (*), 10^k -1 divides 99, implying k =1 or 2. We are done.
A: There is only one prime, 101, in the list.
( A crisp proof comes only after a great deal of scratch work, which is seldom discolsed).

Q) Moscow.
p,q are twin primes, ie, p,q differ by 2. Prove that
p+q | p^q+q^p.
Disc: Without losing generality, let p < q. Note that p is odd, so
p+q | p^p+ q^p.Now
p^q+q^p
= (p^p+q^p)+(p^q- p^p).
The 1st bracket is divisible by p+q. What about the 2nd?
p^q-p^p= p^p(p^2-1), ?
=p^p(p+1)(p-1)
=p^p(p+1)(2n),
since p-1 is even.
=(p^p)n(2p+2)
=(p^p)n(p+q), so the 2nd bracket, too, is divisible by p+q.
The result holds for any two successive odd numbers, p & q.

Another tricky problem:
There are 7 steps in a flight of stairs(not counting the top and bottom of the flight). When going down ,you can jump over some steps,perhaps even over all 7.how many ways are there to go down stairs?

Simple but tricky puzzle to solve.. dm your answers

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