I simply thought that if every odd numbers which space non prime, compulsorily multiples that 3,I tried to gain some starrkingschool.netematics yet in vain. Kindly aid me.

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.