You are watching: Odd numbers that are not prime

No. Because that example, $25 = 5 cdot 5$ or $35 = 5 cdot 7$ space odd, not prime, and not multiples that 3.

More generally, take the product that at the very least two strange primes $ eq 3.$

No. Just take

$X = displaystyle prod_1^n p_i^a_i, ag 1$

where

$2, 3 e p_i in Bbb P, ; a_i in Bbb N, 1 le i le n; ag 2$

that is, $X in Bbb N$ is the product that $n$ prime powers, whereby no prime in the product is $2$ or $3$; climate it is clear that $X$ is odd, because it is a product the odds, and

$3 ot mid X. ag 3$

In this way, a myriad of instances my it is in constructed, *viz.*

$X = 5 cdot 7 cdot 11 = 385; ; X = 11 cdot 13 cdot 17 = 2431, ; X = 29 cdot 31 cdot 37 = 33263, ; extetc. Etc. Etc; ag 4$

the perform of together $X$ is in fact quite lengthy . . .

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answered january 27 "19 in ~ 0:53

Robert LewisRobert Lewis

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The many relevant OEIS entry to your inquiry is most likely A038509, composite number congruent come $pm 1 mod 6$.

Given any type of nonzero integer $k$, that is evident that $6k$ is composite (let"s overlook 0 for now). If $k eq -1$ or 0, climate $6k + 3$ is one odd number, composite and clearly divisible by 3, together you currently know. $6k + 2$ and $6k + 4$ room obviously even however neither is divisible through 3.

So the leaves united state $6k + 1$ and $6k + 5$. Neither is divisible through 2 or 3. They might both be prime. But you should additionally know that the primes thin out together you go further out towards infinity.

In fact, offered a positive integer $n$, friend can constantly find $n$ continually integers such that none of them are prime.

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Let"s try $n = 7$. Collection $k = 15$, then clearly $6k = 90$ is composite, ~ above account of gift divisible through 2, 3 and 5. Therefore 91 can"t it is in divisible by 2, 3 or 5, it could even it is in prime. Nope: $91 = 7 imes 13$. Then 93 is clearly divisible through 3. And 95 is divisible by 5 yet not by 2 or 3, and also it"s obviously no prime.