Concatenation: 2^n || 2^(n+1)

 
 
Let   n   be any positive integer.

2^n \; || \; 2^{n+1}

2^0 \; || \; 2^1 \; = \; 1 \; || \; 2 \; = \; 12 \; = \; 3 \; \times \; 4
2^1 \; || \; 2^2 \; = \; 2 \; || \; 4 \; = \; 24 \; = \; 3 \; \times \; 8
2^2 \; || \; 2^3 \; = \; 4 \; || \; 8 \; = \; 48 \; = \; 3 \; \times \; 16
2^3 \; || \; 2^4 \; = \; 8 \; || \; 16 \; = \; 816 \; = \; 3 \; \times \; 272
2^4 \; || \; 2^5 \; = \; 16 \; || \; 32 \; = \; 1632 \; = \; 3 \; \times \; 544
2^5 \; || \; 2^6 \; = \; 32 \; || \; 64 \; = \; 3264 \; = \; 3 \; \times \; 1088
2^6 \; || \; 2^7 \; = \; 64 \; || \; 128 \; = \; 64128 \; = \; 3 \; \times \; 21376
2^7 \; || \; 2^8 \; = \; 128 \; || \; 256 \; = \; 128256 \; = \; 3 \; \times \; 42752
2^8 \; || \; 2^9 \; = \; 256 \; || \; 512 \; = \; 256512 \; = \; 3 \; \times \; 85504
2^9 \; || \; 2^{10} \; = \; 512 \; || \; 1024 \; = \; 5121024 \; = \; 3 \; \times \; 1707008

 
Prove that   2^n \; || \; 2^{n+1}   is always a multiple of 3.

 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
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