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What is the smallest positive integer, N, that will satisfy the following congruences: N mod (3, 5, 7, 11, 13, 17, 19) = Respectively(1, 2, 3, 4, 5, 6, 7, as remainders)?
Any help or hint would be greatly appreciated. Thank you.

 Oct 3, 2019
 #2
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I assume the question to mean:

 

N mod 3 = 1

N mod 5 = 2

N mod 7 = 3

N mod 11 =4

N mod 13 =5

N mod 17 =6

N mod 19 =7

 

I have used the "Chinese Remainder Theorem + Modular multiplicative Inverse", which are incorporated in this computer code to solve your congruences:

 

i=0;j=0;m=0;t=0;a=(3, 5, 7, 11, 13, 17, 19);r= (1, 2, 3, 4, 5, 6, 7);c=lcm(a); d=c / a[i];n=d % a[i] ;loop1:m++; if(n*m % a[i] ==1, goto loop, goto loop1);loop:s=(c/a[i]*r[j]*m);i++;j++;t=t+s;m=0;if(i< count a, goto4,m=m);printc,"m + ",t % c;return

 

N = 4,849,845 m +  4,383,592, where m =0, 1, 2, 3.........etc.

 Oct 4, 2019

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