Prime p > 3, k = floor(2*p/3), c(p,1)+c(p,2)+c(p,3)+…+c(p,k)

 
 

Prove that for all   p   prime number   (   p > 3  )   and   k \; = \; \lfloor \, (2 \, p/3) \,\rfloor

then   \dbinom{  \:p  \:}{  \:1  \:} \; + \; \dbinom{  \:p  \:}{  \:2  \:} \; + \; \dbinom{  \:p  \:}{  \:3  \:} \; + \; ... \; + \; \dbinom{  \:p  \:}{  \:k  \:}   is divisible by   p^2

 
For example,
 

p \; = \; 5

\lfloor \, (2\times5/3) \,\rfloor \; = \; 3

\dbinom{  \:5  \:}{  \:1  \:} \; + \; \dbinom{  \:5  \:}{  \:2  \:} \; + \; \dbinom{  \:5  \:}{  \:3  \:} \; = \; 25 \; = \; 5^2

 

p \; = \; 7

\lfloor \, (2\times7/3) \,\rfloor \; = \; 4

\dbinom{  \:7  \:}{  \:1  \:} \; + \; \dbinom{  \:7  \:}{  \:2  \:} \; + \; \dbinom{  \:7  \:}{  \:3  \:} \; + \; \dbinom{  \:7  \:}{  \:4  \:} \; = \; 98 \; = \; 2 \times 7^2

 

p \; = \; 11

\lfloor \, (2\times11/3) \,\rfloor \; = \; 7

\dbinom{  \:11  \:}{  \:1  \:} \; + \; \dbinom{  \:11  \:}{  \:2  \:} \; + \; \dbinom{  \:11  \:}{  \:3  \:} \; + \; ... \; + \; \dbinom{  \:11  \:}{  \:7  \:} \; = \; 1815 \; = \; 3 \times 5 \times 11^2

 

p \; = \; 13

\lfloor \, (2\times13/3) \,\rfloor \; = \; 8

\dbinom{  \:13  \:}{  \:1  \:} \; + \; \dbinom{  \:13  \:}{  \:2  \:} \; + \; \dbinom{  \:13  \:}{  \:3  \:} \; + \; ... \; + \; \dbinom{  \:13  \:}{  \:8  \:} \; = \; 7098 \; = \; 2 \times 3 \times 7 \times 13^2

 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

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