# Is there a prime number $> 10$ such that when it is divided by 3 or 5 or 7 always gives a remainder of 1?

Is there a prime number $p > 10$ such that when it is divided by 3 or 5 or 7 gives a remainder of 1, i.e.:

$p \equiv 1 \pmod{3}, p \equiv 1 \pmod{5}, p \equiv 1 \pmod{7}$.

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Yes: look up Dirichlet's theorem on primes in arithmetic progression. – Geoff Robinson Dec 31 '12 at 17:48
I guess that there is other post here with the same questions but I'm not able to find it. – Sigur Dec 31 '12 at 17:48
To expand Geoff's comment: There are infinitely many primes satisfying these congruences. – Andrés Caicedo Dec 31 '12 at 17:54

$$x \equiv 1 \pmod{3}$$ $$x \equiv 1 \pmod{5}$$ $$x \equiv 1 \pmod{7}$$ Hence, $$x \equiv 1 \pmod{105}$$ Since we want $x$ to be a prime, we have $$x \equiv 1 \pmod{210}$$ $x=211$ happens to be a prime.

As an aside and completely irrelevant to the post, one of G H Hardy's desire was to make a match-winning score of $211$ not-out in cricket in the last innings at Oval. (since $211$ is the first prime after $200$ and G H Hardy had a great passion for cricket. Oval is one of the most famous cricket ground in England.)

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Nice aside, Marvis - interesting piece of math trivia. – Conan Wong Dec 31 '12 at 17:59
How many prime numbers like this are out there? any idea or we just need to multiple it by 2 (421) – Shan Dec 31 '12 at 18:02
@Shan By Dirichlet's theorem on primes, there are infinitely many primes. As a crude way of saying, $\dfrac1{80}$ of all the primes are of this form, the number $80$ appearing in the denominator since $80 = \phi(220)$ i.e. there are $80$ numbers less than $220$ and relatively prime to it. – user17762 Dec 31 '12 at 18:05
I do know about Oval :) – Shan Dec 31 '12 at 18:20

By the Chinese Remainder Theorem, the system of linear congruences:

$$x \equiv 1 (mod \ 3)$$ $$x \equiv 1 (mod \ 5)$$ $$x \equiv 1 (mod \ 7)$$

yields the solution $x \equiv 1 (mod \ 105)$, since 3, 5 and 7 are pairwise coprime and their product is 105.

Since we are looking for positive integers, we consider $x = 1, 106, 211,...$ and we need not look further since 211 is a prime.

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+1 for mentioning the Chinese Remainder Theorem. – KeithS Dec 31 '12 at 19:59
Thanks Keith - appreciate it. – Conan Wong Dec 31 '12 at 20:34
TeX note: if you use \pmod{3} instead of  (mod 3) the upright font will be used instead of the italics which makes it look like the product $mod=m\cdot o\cdot d$ or the composition $m\circ d$. – Charles Jan 2 '13 at 16:51