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# help is greatly appreciated thank youu

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Find all numbers r for which the system of congruences

x = r (mod 6),

x = 9 (mod 20),

x = 4 (mod 45)

has a solution.

*Sorry I didn't know how to put the sign with 3 straight horizontal lines so I just made them equal signs.

Guest Aug 3, 2017
edited by Guest  Aug 3, 2017
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#1
+1

x mod 20=9, x mod 45=4, x mod 6 =r, solve for x, r

By simple iteration, the smallest positive integer that satisfies all the congruences is=49

Check:

49 / 20 = 2 with a remainder of 9, and:

49 /45 =1  with a remainder of 4, and:

49 / 6 =8   with a remainder of 1. Therefore:

r=1 and x=49

Guest Aug 3, 2017
#2
+1

You can extend Guest#1 answer to include the following:

Since the LCM of: 6, 45, 20 =180, therefore:

x =180n + 49, where n=0, 1,2,3.....etc.

So that: x=49, 229, 409, 589.....etc.

Guest Aug 3, 2017

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