sum of 3 out of 5 is divisible by 3

Problem:
Among 5 integers, there are always 3 with sum divisible by 3.
(From Problem Solving Strategies, Arthur Engel)

Solution:
Proof by Pigeon Hole Principle.
An integer is either 0, 1 or 2 \( \pmod 3 \). Imagine placing 5 integers into those 3 boxes. If we have at least one in each, then we can pick one from each, with sum divisible by 3.
Otherwise, we'll have at least 3 integers in one of the boxes. Pick those 3. QED

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

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 ...

n-th non-square

Problem: Find the n-th non-square number. (Adapted from Problem Solving Strategies, Arthur Engel ) If you have CS background, I invite you to come up with an O(log N) solution. Done? Good. Below I present the O(1) approach. Solution: The n-th non-square number is given by \( F(n) = \left \lfloor n + \sqrt{n} + 1/2 \right \rfloor \). Proof: Let G(n) be the same as F(n) except for the floor function. Investigate \( m \neq F(n) \). There is the largest \( n \) s.t. \( F(n) < m \). Hence \( m+1 \leq F(n+1) \). We conclude that \(G(n) < m \) and \(m+1 \leq G(n+1) \). Rearranging the terms and squaring the inequalities, we get: \( n < (m-n-1/2)^2 \) and \( (m-n-1/2)^2 < n+1 \), or \( n - 1/4 < (m-n)^2 - (m-n) < n + 3/4 \). Since the middle term is an integer, we see that \( (m-n)^2 - (m-n) = n \). Simplifying, we get \(m = (m-n)^2\), showing that m is a square. In other words, F(n) skips all squares. Now, we will show that F(n) is the n-th non...

Maths by Chance

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