Proving Every Int. ≥ 12 Is a Combination of 4m + 5n

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To prove that every integer greater than or equal to 12 can be expressed as 4m + 5n, where m and n are non-negative integers, mathematical induction is suggested as the method of proof. The base case is established at n=12, confirming that it can be represented as 4(3) + 5(0). The discussion hints at using the relationship 1 = 5 - 4 to facilitate the induction step. Participants emphasize the importance of demonstrating that if the statement holds for an integer k, it must also hold for k+1. This approach ultimately aims to show that all integers from 12 onward can be formed using the specified combination of 4 and 5.
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Homework Statement



Prove that every integer greater than or equal to 12 can be written as a combination of 4m+5n where m and n are non-negative integers.


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The Attempt at a Solution



I know I have to use induction but I don't really know how to go about doing it past showing that it works for the base case of n=12 where then a=3 and b=0.
 
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Hint: 1 = 5 - 4
 

Thread 'Family of lines that are at a distance of 5 from the origin'

I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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