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Find an integer a > 2 such that 2|a , 3|(a+1) , 4|(a+2), 5|(a+3), and 6|(a+4)

Using the definition of congruence 2|a implies that a congruent 0 mod 2

because a congruent b mod n implies that m|(a-b). My idea is to set up a system of linear congruences and apply the Chinese remainder theorem.. but I can't figure out how to put 3|(a+1) 4|(a+2), 5|(a+3), and 6|(a+4) into congruence statements. My notes say 3|(a+1) implies a congruent 2 mod 3 but I am not sure why this is true.. any help?

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    $a+1\equiv 0\pmod{3}\implies a\equiv -1\pmod{3}$ – 2017-02-22
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    As a hint for the solution, note that if $a$ passes this test, then so does $a+(3\times 4\times 5)n$ – 2017-02-22
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    Note that of $2\mid a$, $3\mid (a+1)$ and $6\mid (a+4)$, any two implies the third, so we may as well discard the last one. Also, $4\mid (a+2)$ implies $2\mid a$, so we can discard that one too. We left with $3\mid(a+1),4\mid(a+2),5\mid(a+3)$. That might be a bit more manageable. – 2017-02-22

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