Read the proof below. If the numbers given in the second sentence are {42, 43, 55, 74, 96, 127, 130}, then which boxes end up with at least two num
Read the proof below. If the numbers given in the second sentence are {42, 43, 55, 74, 96, 127, 130}, then which boxes end up with at least two num
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
Please explain.

Transcribed Image Text:**Understanding the Proof: Divisibility by 10**
**Problem Statement:**
Given the numbers {42, 43, 55, 74, 96, 127, 130}, identify which boxes will end up with at least two numbers. Select all applicable options.
**Interactive Response:**
The correct answer is Box 3 and Box 4.
**Claim:**
Among any 7 integers, there are two whose sum or difference is divisible by 10.
**Proof:**
We categorize numbers into six boxes based on their last digit:
| Box Number | Last Digit |
|------------|-------------|
| 0 | 0 |
| 1 | 1 or 9 |
| 2 | 2 or 8 |
| 3 | 3 or 7 |
| 4 | 4 or 6 |
| 5 | 5 |
**Explanation:**
When you place any 7 integers into these six categories, the Pigeonhole Principle dictates that at least one box will have two or more integers. If two integers share the same last digit, their difference is divisible by 10. If different, based on the box definitions, their sum will be divisible by 10.
Expert Solution

Step 1
According to given box model,
Box number | 0 | 1 | 2 | 3 | 4 | 5 |
Last digit | 0 | 1 or 9 | 2 or 8 | 3 or 7 | 4 or 6 | 5 |
Given numbers are,
42, 43, 55, 74, 96, 127, 130.
Step by step
Solved in 2 steps
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