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Jan 20, 2019
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Jan 20, 2019
 #2
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There is a more formal method to solving this problem.

First let's reduce these congruences.

 

7x21(mod14)x3(mod2)x1(mod2)

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2x+1316(mod9)2x3(mod9)now we multiply both sides by the multiplicative inverse of 2(mod9)52x53(mod9)x6(mod9)

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2x+1x(mod25)3x1(mod25)173x17(mod25)x17(mod25)

 

Now we have the following system of linear congruencesx1(mod2)x6(mod9)x17(mod25)

 

From the first congruence we havex=2t+1, tZsubstitute this into the second congruence2t+16(mod9)2t5(mod9)52t55(mod9)t7(mod9)t=9s+7, sZ

 

substitute this back into x and simplifyx=2(9s+7)+1x=18s+14+1=18s+15

 

Now we substitute this into the third congruence18s+1517(mod25)18s2(mod25)Now we have to find 181(mod25), with these small numbers trial and error works181(mod25)=7718s14(mod25)s14(mod25)s=25u+14, uZ

 

and we substitute this back into xx=18(25u+14)+15=267+450ux267(mod450)The smallest 4 digit solution will bex=2450+267=1167

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Jan 20, 2019

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