Category: Math

A Prime Factorisation Case

24 Jul 18
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My dear friend Srinivas Shastri once asked this question in some public forum. I had a lovely time solving this with my son. It was a simple problem with a simpler solution. Unfortunately a very small fraction of the people I have asked have been able to solve this one.

The Problem?

Basically, you take any prime number greater than 3. Say 5 for starters.

You square it. i.e. you get 25

You subtract 1 giving 24.

Now, take a bigger prime number, let us say 23 and run the same operations on it.

Squaring it gives 529. Subtracting 1 gives 528. This is equal to 22 times 24! Is this coincidence???

Take an even bigger number. 43 (not too big because we need to be able to do it mentally for speed)

Squaring it gives 1849. Subtracting 1 gives 1848. This is equal to 77 times 24. Have we proved it by induction?

Now, how do we go about proving this generically?

The approach to the solution

We all know that

  • Now, since n has to be odd, (n+1) and (n-1) are both even.
  • Given any two consecutive even numbers, one is divisible by 2 and the other by 4
  • Further, in any set of 3 consecutive numbers (n-1), n and (n+1), one of them has to be divisible by 3
  • Since it cannot be n, it is either (n-1) or (n+1)
  • From the statements above, it is clear that 2,3 and 4 are all factors of the number we are looking to divide.
  • And the product of these numbers is 24!!!!

Such a simple, elegant proof!.

Which is Bigger?

23 May 18
Sukrit Vijaykar
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Sum fun things about numbers courtesy my friend Srinivas Shastri a very wise lover of maths

There was this question once asked in JEE

Which is greater? e^π or π^e?

The solution to this problem is simple yet elegant

The trick is to see that the function x^(1/x) is max when x=e. So:

e^(1/e) > π^(1/π)

Raising both sides to eπ, we get:

From our IIT-JEE 1981 Math paper
with a doff to Feynman.QED



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