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# remainder when x is divided by 105

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Suppose that x is an integer that satisfies the following congruences:

3 + x == 2^2 (mod 3^3)

5 + x == 3^2 (mod 5^3)

7 + x == 4^2 (mod 7^3)

What is the remainder when x is divided by 105?

Oct 22, 2020

#1
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Using CRT and MMI:

i=0;j=0;m=0;t=0;a=(27, 125, 343);r= (1, 4, 9);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

LCM(3^3, 5^3, 7^3) =1,157,625

x = 1157625 m +  759754, where m = 0, 1, 2, 3, ........etc.

The smallest x =759,754 mod 105 = 79 - the remainder.

Oct 23, 2020
#2
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Suppose that x is an integer that satisfies the following congruences:

3 + x == 2^2 (mod 3^3)

5 + x == 3^2 (mod 5^3)

7 + x == 4^2 (mod 7^3)

What is the remainder when x is divided by 105?

3 + x == 2^2 (mod 3^3)

3 + x == 4 (mod 3^3)

x == 1 (mod 3^3)

x == 1 (mod 3)

5 + x == 3^2 (mod 5^3)

x == 4 (mod 5^3)

x == 4 (mod 5)

x=-1 (mod 5)

7 + x == 4^2 (mod 7^3)

x == 9 (mod 7^3)

x == 9 (mod 7)

x == 2 (mod 7)

Look for a number that works here.

9,16,23,30,37,44,51,58,65,72,79

x could be 79 except this it too little for the original mods.

3*5*7=105

So

x=79+105K     Where k is a positive integer  (not all positive integers will work)

So    x==79  (mod105)

Oct 23, 2020
edited by Melody  Oct 23, 2020