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KCC's Quiz AQQ300 about Divisibility by 6

		
	
				

			

									
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  •  Analog Employees  

			
			
			    	
			                    
                
				                on May 8, 2026 
				
			
			
		

		
		                     
				
        		

		    
		    		    
		    			

			    
1. Quote of the month: "Friendship is like money - easier made than kept" - Samuel Butler




2. New quiz AQQ300 about a divisibility by 6 puzzle 


 


Good luck and try to be among the first ones!



			

			
									
		                             
                        

    

                                            
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    •  Analog Employees  
    
            
						
                            
            
            				
            
                on May 12, 2026 4:52 PM
            
            
						
			            		
    
		
								
	
    

		
		
    
		
    
                                
                                                            
    
					
    

    (n5−n) is always divisible by 6. We will prove this by showing that it is always divisible by both 2 and 3.

    

    (n^5 - n) is always divisible by 2 because it is always even:

     if n is even, then (n^5 - n) is even because it is the difference of two even numbers;

     if n is odd, then n^5 is odd, therefore (n^5 - n) is the difference of two odd numbers, and hence it is an even number.

    

    We now prove that (n^5−n) is always divisible by 3.

    

    (n^5−n)=n(n^4−1)=n(n^2−1)(n^2+1)=n(n−1)(n+1)(n^2+1).

    

    If n=3k, then (n^5−n) is obviously divisible by 3.

     If n is not divisible by 3, then one of the terms (n - 1) or (n+1) certainly is, because for every natural integer n only the following cases are possible:

    

    n=3k, n=3k+1, n=3k+2
    

    

    Therefore, (n^5−n) is always divisible by both 2 and 3, and hence by 6.
    

    
			
    

		
			
	      

		
		
	
    
	
    
		
    
			
		
    
	
    
	
    
		
						
				
				
						
				
				
		
		
    
			
		
    
	
    

    

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    •  Analog Employees  
    
            
						
                            
            
            				
            
                on May 12, 2026 4:52 PM
            
            
						
			            		
    
		
								
	
    

		
		
    
		
    
                                
                                                            
    
					
    

    (n5−n) is always divisible by 6. We will prove this by showing that it is always divisible by both 2 and 3.

    

    (n^5 - n) is always divisible by 2 because it is always even:

     if n is even, then (n^5 - n) is even because it is the difference of two even numbers;

     if n is odd, then n^5 is odd, therefore (n^5 - n) is the difference of two odd numbers, and hence it is an even number.

    

    We now prove that (n^5−n) is always divisible by 3.

    

    (n^5−n)=n(n^4−1)=n(n^2−1)(n^2+1)=n(n−1)(n+1)(n^2+1).

    

    If n=3k, then (n^5−n) is obviously divisible by 3.

     If n is not divisible by 3, then one of the terms (n - 1) or (n+1) certainly is, because for every natural integer n only the following cases are possible:

    

    n=3k, n=3k+1, n=3k+2
    

    

    Therefore, (n^5−n) is always divisible by both 2 and 3, and hence by 6.
    

    
			
    

		
			
	      

		
		
	
    
	
    
		
    
			
		
    
	
    
	
    
		
						
				
				
						
				
				
		
		
    
			
		
    
	
    

    

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