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Determine the smallest non-negative integer  that satisfies the congruences:

\(\begin{align*} &a\equiv 2\pmod 3,\\ &a\equiv 4\pmod 5,\\ &a\equiv 6\pmod 7,\\ &a\equiv 8\pmod 9. \end{align*}\)

RektTheNoob  Jul 30, 2017
 #1
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Since the LCM of  3, 5, 7, 9 =315, and the remainders are 1 less than the moduli in each case, therefore the smallest number that satisfies the congruences is:

315 - 1 = 314, so that you have:

315N + 314 and N =0, 1, 2, 3......etc. So:

a = 314, 629, 944.....etc. 

Guest Jul 30, 2017
edited by Guest  Jul 30, 2017
 #3
avatar+93903 
+1

Thanks Guest, I just wanted to think this one through myself :)

 

a=9k1-1    that satisifes both the first and the last congruences

a=5k2-1

a=7k3-1

 

So if a is 1 less than a multiple of 5 and 9 and 7 then it must be 1 less than any common factor of 9 and 5 and 7


So the smallest one is 1 less than 315

 

answer 314

Melody  Jul 31, 2017

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