Skip to content


Two Simple Math Puzzles: Prime Numbers and Shortest Path

Posted on:June 20, 2015 at 12:00 AM
 at 2 min read

Here are two math puzzles, solve, comment and enjoy the discussion!

Puzzle 1: Prime NumbersSection titled Puzzle 1: Prime Numbers

Prove that p21p^2-1 is divisible by 24, where pp is a prime number with p>3p>3.

This is a simple one - do not go by the technicality of the problem statement.

Solution 1Section titled Solution 1

p21=(p1)×(p+1)p^2-1 = (p-1)\times (p+1) Given, pp is a prime number >3>3, p1p-1 and p+1p+1 are even. We can also write,

p1=2K,and p+1=2K+2=2(K+1)\begin{aligned} p-1=2K, \text{and } \cr p+1=2K+2=2(K+1) \end{aligned}

Given, KNK \in \mathbb{N}, either KK or K+1K+1 are also even. Hence, (p1)×(p+1)(p-1)\times (p+1) is divisible by 2×4=82\times 4 = 8.

Furthermore, as pp is prime, we can write it as either p=3s+1p = 3s+1 or p=3s1p = 3s-1, where sNs \in \mathbb{N}. In either case, one of p1p-1 or p+1p+1 are divisible by 3 as well.

Hence, p21p^2-1 is divisible by 8×3=248\times 3 = 24.

Puzzle 2: Hopping on a Chess BoardSection titled Puzzle 2: Hopping on a Chess Board

On a chess board, basically a 8×88\times 8 grid, how many ways one can go from the left most bottom corner to the right most upper corner? In chess naming conventions, from a1a1 to h8h8.

chess board

Can you find the generic answer for the case of N×MN\times M grid?

Solution 2Section titled Solution 2

Given only forward moves are allowed, for any arbitrary grid of N×MN\times M, a total of (N1)+(M1)(N-1) + (M-1) moves are needed.

Any N1N-1 of these moves can be of type right and M1M-1 of type up. In short, this is a combinatorial problem of distributing N+M2N+M-2 objects into groups of N1N-1 and M1M-1. This is simply,

(N+M2N1)=(N+M2)!(N1)!(M1)!\dbinom{N+M-2}{N-1} = \frac{(N+M-2)!}{(N-1)! (M-1)!}

In the particular case of the chess board, N=M=8N = M = 8. Hence, total number of possible paths are:

No. of Paths=14!7!7!=3432\text{No. of Paths} = \frac{14!}{7! 7!} = 3432

Let me know if you have any other interesting alternative solutions to these problems.


RELATED POSTS | Algorithms