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

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section: Chapter Questions
Problem 64RE
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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 numbers? (Select all)
Box 0 Box 1 Box 2 Box 3 Box 4 Box 5
Claim. Among any 7 integers, there are two whose sum or difference is divisible by 10.
Proof. We define six boxes, so that each integer is placed among them based on the integer's
last digit:
Box number 0 1
Last digit
0 1 or 9
Box 3
Box 4
2
2 or 8
3
3 or 7
4
5
4 or 6 5
Now given any 7 integers, when placed among these 6 boxes, some box will contain at least two
integers by the Pigeonhole Principle. If these two integers have the same last digit, then their
difference will be divisible by 10. Otherwise, by the definition of the boxes, the two numbers
will have a sum that is divisible by 10.
Transcribed Image Text: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 numbers? (Select all) Box 0 Box 1 Box 2 Box 3 Box 4 Box 5 Claim. Among any 7 integers, there are two whose sum or difference is divisible by 10. Proof. We define six boxes, so that each integer is placed among them based on the integer's last digit: Box number 0 1 Last digit 0 1 or 9 Box 3 Box 4 2 2 or 8 3 3 or 7 4 5 4 or 6 5 Now given any 7 integers, when placed among these 6 boxes, some box will contain at least two integers by the Pigeonhole Principle. If these two integers have the same last digit, then their difference will be divisible by 10. Otherwise, by the definition of the boxes, the two numbers will have a sum that is divisible by 10.
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