# Is it true that all senary numbers ending in 1 and 5 are primes?

I was reading the Wikipedia article on senary numbers (base 6), which states that:

all primes, when expressed in base-six, other than 2 and 3 have 1 or 5 as the final digit

Unless I am converting to senary incorrectly, I find this not to be true. For example, the senary representation of the decimal number 2047 is '13251', which would be a prime according to the stated rule, but is not (2047 = 89 * 23).

Is my conversion correct? Is the stated rule incorrect?

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I think the statement says that all primes other than $2$ or $3$ have $1$ or $5$ as the final digit, not that all numbers that have $1$ or $5$ in base $6$ as the final digit are primes. So I believe your example shows that the converse is not true, but this is not the same as saying the statement is not true. –  yunone Mar 19 '11 at 0:19
In other words some non-primes may also end in 1 or 5... Ahhh. Too obvious to see. Thanks... Secondary question, then: does anyone know if numbers in base 6 which end and 1 or 5 that are not primes have anything in common? –  Benjamin Mar 19 '11 at 0:24
What they all have in common is that they are not divisible by $2$ or $3$. There isn't much else to be said, because having a base $6$ expansion that ends in $1$ or $5$ is equivalent to not being divisible by $2$ or $3$ –  Jonas Meyer Mar 19 '11 at 0:29
In particular, the smallest example is $5^2=41_6$, and the next smallest is $5\cdot 7=55_6$. –  Jonas Meyer Mar 19 '11 at 0:39
I don't think this question should be closed. It contains two pieces of interest to mathematicians: senary numbers and mathematical logic (misunderstanding $\,A\implies B\,$ as meaning (that also) $\,B\implies A\,$ ) –  DonAntonio May 16 '13 at 12:01

You are misinterpreting the statement. "All primes satisfy property $X$" means "If $p$ is prime, then $p$ has property $X$." You have instead interpreted it as "If $p$ has property $X$, then $p$ is prime."

The statement is true, because if $p$ is a prime greater than $3$, then $p$ is not divisible by $2$ or $3$, whereas a number whose base six expansion ends in $0$, $2$, or $4$ is even and a number whose base six expansion ends in $0$ or $3$ is a multiple of $3$.

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As said, you have misread the statement. This is a special case of the fact that for an odd prime $\rm\:p\:,\:$ $\rm\: \phi(2\:p)\ =\ \phi(p)\ =\ p-1\:,\$ i.e. there are $\rm\: p-1\:$ naturals below $\rm\:2\:p\:$ that are coprime to $\rm\:2\:p\:,\:$ namely all $\rm\:p\:$ odd numbers below $\rm\:2\:p\$ excepting $\rm\:p\:.\$ Hence, modulo $\rm\:2\:p\:,\:$ an odd prime $\rm\ne p\:$ must lie in one of these congruence classes (else it has a nontrivial gcd with $\rm\:2\:p\:,\:$ so it is composite). $\:$ Hence if $\rm\:q\:$ is prime then $\rm\ q\equiv 1,5\ \ (mod\ 6)\:;\ \ q\equiv 1,3,7,9\ \ (mod\ 10)\:;\ \ q\equiv 1,3,5,9,11,13\ \ (mod\ 14)\$ etc, assuming that $\rm\:q\:$ is coprime to the modulus. Exploiting reflection symmetry we can state this more succinctly: $\rm\ \ q\equiv \pm 1\ \ (mod\ 6)\:;\ \ q\equiv \pm\{1,3\}\ \ (mod\ 10)\:;\ \ q\equiv\pm \{1,3,5\}\ \ (mod\ 14)\$ and, more generally, $\rm\:\ \ \ q\equiv \pm\{1,3,5,\cdots,p-2\}\ \ (mod\ 2\:p)\$

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The relationship in the quote is a subset. All primes in base 6, other than 2 and 3, is a subset of all numbers ending in 1, 5. There are composite numbers that end in the same digits, since all numbers coprime to six end in 1,5 base 6.

41 = dec 25 = 5*5 , and 205 = 11 * 15 = dec 77 evidently are not primes, but are co-prime to 6.

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