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Find the least positive integer x that satisfies \(x+4609 \equiv 2104 \pmod{12}\).

tertre  Mar 23, 2017
 #1
avatar
0

The smallest positive integer =3

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

Guest Mar 23, 2017
 #2
avatar+7023 
+1

if \(x+4609 \equiv 2104 \pmod{12}\) , 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.

MaxWong  Mar 23, 2017
 #3
avatar+20009 
+1

Find the least positive integer x that satisfies 

\(x+4609 \equiv 2104 \pmod{12}\).

 

\(\begin{array}{|lrcll|} \hline & x+4609 &\equiv& 2104 \pmod{12} \\ & x+4609-2104 &=& n \cdot 12 \quad & | \quad n \in \mathbb{N} \\ & x+2505 &=& n \cdot 12 \\ & x&=& n \cdot 12 -2505 \\\\ \text{so } & n \cdot 12 &=& 2505 \\ & n &=& \uparrow \frac{2505}{12} \uparrow \\ & n &= & \uparrow 208.75 \uparrow \\ & n &= & 209 \\\\ & x&=& 209 \cdot 12 -2505 \\ & x&=& 2508 -2505 \\ & \mathbf{x} & \mathbf{=} & \mathbf{3} \\ \hline \end{array}\)

 

laugh

heureka  Mar 24, 2017

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