Is there any elegant way to prove that 28C14 -1 is divisible by 29? Also, is this kind of a result a theorem or a generalisation? If so please do help...
note i do mean 28 choose 14
2026-03-27 12:03:06.1774612986
Looking for divisibility by 29 and a general proof( if any)
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I presume you mean $\binom{28}{14}-1$?
Note that $$14!\binom{28}{14}=28\times27\times26\times\cdots\times 15 \equiv(-1)(-2)(-3)\cdots(-14)=14!\pmod{29}.$$ Therefore $\binom{28}{14}\equiv1\pmod{29}$.