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

 

a = 2 (mod 3)

a = 4 (mod 5)

a = 1 (mod 7)

a = 8 (mod 9)

 Jul 3, 2021
 #1
avatar+2407 
+1

If a = 8 (mod 9), then a must be 2 (mod 3), so we can forget about that equation. 

 

a = 4 (mod 5)

a = 1 (mod 7)

a = 8 (mod 9)

 

I'm going to use the Chinese Remainder Theorem, which is a bit difficult to exaplain, but there are some good videos on youtube. 

 

Starting with mod 5. 

x = 0 (mod 7) and (mod 9) but 4 (mod 5). 

x = 7*9*n = 4 (mod 5)

This works when n = 3, x = 189. 

 

Then, mod 7. 

x = 0 (mod 5) and (mod 9) but 1 (mod 7). 

x = 5*9*n = 1 (mod 7).

This works when n = 5, x = 225. 

 

Then, mod 9. 

x = 0 (mod 5) and (mod 7) but 8 (mod 9). 

x = 5*7*n = 8 (mod 9).

This works when n = 1, x = 35. 

 

Finally, we add everything up. 

189 + 225 + 35 = 449

We can make it smaller by doing 449 - 5*7*9 = 134. 

 

=^._.^=

 Jul 3, 2021
 #2
avatar+26396 
+3

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

a2(mod3)a4(mod5)a1(mod7)a8(mod9)

 

1. join
a2(mod3)a8(mod9)

a=2+3na=8+9m2+3n=8+9m3n=6+9m|:3n=2+3ma=2+3n|n=2+3ma=2+3(2+3m)a=2+6+9ma=8+9ma8(mod9)a2(mod3)a8(mod9)}a8(mod9)

 

2. join
a4(mod5)a1(mod7)

a=4+5ra=1+7s4+5r=1+7s5r=7s3r=7s35r=5s+2s35r=s+2s35=tr=s+tt=2s355t=2s32s=5t+3s=5t+32s=4t+t+2+12s=2t+1+t+12=us=2t+1+uu=t+122u=t+1t=2u1s=2(2u1)+1+us=5u1r=5u1+2u1r=7u2a=4+5ra=4+5(7u2)a=6+35ua6(mod35)a4(mod5)a1(mod7)}a6(mod35)

 

3. join
a8(mod9)a6(mod35)

a=8+9va=6+35w8+9v=6+35w9v=35w14v=35w149v=27w+8w959v=3w1+8w59=xv=3w1+xx=8w599x=8w58w=9x+5w=9x+58w=8x+x+58w=x+x+58=yw=x+yy=x+588y=x+5x=8y5w=8y5+yw=9y5v=3(9y5)1+8y5v=35y21a=8+9va=8+9(35y21)a=181+315ya181(mod315)a181+315(mod315)a134(mod315)a8(mod9)a6(mod35)}a134(mod315)

 

The smallest non-negative integer a is 134

 

laugh

 Jul 4, 2021

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