Knight through the board

Problem:

Starting from the top-left corner of the chessboard, can we move a knight to the bottom-right corner such that every square is visited once?

Solution:

No. Each step, the square the knight is in switches color. There are 63 (i.e. odd) steps to make, hence the last square must be of a different color from that of the starting point. However the top-left corner and the bottom-right corner have the same color. Contradiction.

Comments

641 | 2^32 + 1

Problem: Show that \( 641 | 2^{32} + 1 \). Solution: (From Problem Solving Strategies, Arthur Engel) \( 641 = 625 + 16 = 5^4 + 2^4 \). So \( 641 | 2^{32} + 2^{28} \cdot 5^4 \). Also, \( 641 = 640 + 1 = 2^7 \cdot 5 + 1\). So \( 641 | (2^7 \cdot 5)^4 - 1 = 2^{28}\cdot 5^4 - 1 \). Hence \( 641 | 2^{32} + 2^{28} \cdot 5^4 -(2^{28}\cdot 5^4 - 1) \). QED

2n+1 3n+1 squares then 5n+3 not prime

Problem: Prove that if \( 2n+1 \) and \( 3n+1 \) are both squares, then \( 5n+3 \) is not a prime. Solution: Proof by contradiction. Suppose that there is n such that 5n+3 is prime. Note that \( 4(2n+1) - (3n+1) = 5n+3 \). Let \( 2n+1 = p^2 \) and \( 3n+1 = q^2 \), where \( p, q > 0 \). We have \( (2p-q)(2p+q) = 5n+3 \). Since RHS is a prime, we must have \(2p - q = 1 \) and \( 2p + q = 5n+3 \). Solving for \( q \) we get \( 2q = 5n + 2 \). Substituting, we get \( 2q = q^2 + 2n + 1 \), or \( -2n = (q-1)^2 \). Since RHS is \( \geq 0 \), we can only have equality when \( n= 0 \) and \( q = 1 \). In this case, we have \( 5n + 3 = 8 \) not a prime. QED.

12 coins

Problem: There are 12 identical coins. There might be one that is a counterfeit, which is either heavier or lighter than the rest. Can you identify the counterfeit (if any) using a balance at most 3 times? Solution: I think this is one of the most difficult variations of the coins weighing problem. The main idea is to "mark" the coins after weighing them. Step 1. Divide the coins into groups of 4. Weigh the first two. If they have the same weight, go to step 4. Otherwise, mark the coins belonging to the heavier group 'H', lighter group 'L', and the unweighted ones 'S'. We know that the counterfeit is either in L or H. Step 2. Now we have 4 Hs, 4 Ls, and 4 Ss. We now form 3 groups: HHL, LLH, and HLS. Step 3. Weigh HHL against HLS. Case 3.1: HHL is lighter than HLS. Then either the L in HHL or the H in HLS is the counterfeit. Simply weigh one of them against S and conclude accordingly. Case 3.2: HHL is heavier than HLS. Then the counterfeit ...

Maths by Chance

Search This Blog