Loading [MathJax]/jax/output/SVG/jax.js
 
+0  
 
0
1029
3
avatar+4624 

Find the least positive integer x that satisfies x+46092104(mod12).

 Mar 23, 2017
 #1
avatar
0

The smallest positive integer =3

General formula: 3 + 12n, for n = 0, 1, 2,.....etc.

 Mar 23, 2017
 #2
avatar+9675 
+1

if x+46092104(mod12) , x + 2505 is divisible by 12

x must be an odd number in order to make x + 2505 divisible by 12.

Trial and error:

Try x = 1, 1 + 2505 = 2506 <-- not divisible by 12.

Try x = 3, 3 + 2505 = 2508 <-- divisible by 12(2508 = 2 x 2 x 3 x 11 x 19)

 

Therefore the least positive integer x is 3.

 Mar 23, 2017
 #3
avatar+26396 
+1

Find the least positive integer x that satisfies 

x+46092104(mod12).

 

x+46092104(mod12)x+46092104=n12|nNx+2505=n12x=n122505so n12=2505n=250512n=208.75n=209x=209122505x=25082505x=3

 

laugh

 Mar 24, 2017

2 Online Users