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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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    To expand Geoff's comment: There are infinitely many primes satisfying these congruences.2012-12-31

2 Answers 2

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$x \equiv 1 \pmod{3}$ $x \equiv 1 \pmod{5}$ $x \equiv 1 \pmod{7}$ 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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    I do know about Oval :)2012-12-31
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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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    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 $m\circ d$.2013-01-02