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(A)  Find a number $0\leq x<14$ that solves the congruence $9x \equiv 9 \pmod{14}$.

(b)  Find a number $0\leq x<5$ that solves the congruence $3x \equiv 4 \pmod{5}$.

(c)  Find a number $0\leq x<1000$ that solves the congruence $999x \equiv 998 \pmod{1000}$.

(d)  Find a number $0\leq x<21$ that solves the congruence $4x \equiv 17 \pmod{21}$.

 Jul 30, 2015

Best Answer 

 #1
avatar+26367 
+5

$$\small{
\begin{array}{llrl}
$(A)$ &\text{Find a number }$0\leq x< &14 &$that solves the congruence $9x \equiv 9 \pmod{14}$.\\
$(B)$&\text{Find a number }$0\leq x< &5 &$that solves the congruence $3x \equiv 4 \pmod{5}$.\\
$(C)$&\text{Find a number }$0\leq x< &1000&$that solves the congruence $999x \equiv 998 \pmod{1000}$.\\
$(D)$&\text{Find a number }$0\leq x< &21 &$that solves the congruence $4x \equiv 17 \pmod{21}$.
\end{array}
}$$

 

$$\left\{
\begin{array}{l}
$(A)$\\
$(B)$\\
$(C)$\\
$(D)$
\end{array}
\right\}=
\left\{
\begin{array}{lcr}
x&=&1\\
x&=&3\\
x&=&2\\
x&=&20
\end{array}
\right\}, \small{\text{ because }
\left\{
\begin{array}{lcr}
9\cdot 1 &\equiv& 9 \pmod{14}\\
3\cdot 3 &\equiv& 4 \pmod{5}\\
999\cdot 2 &\equiv& 998 \pmod{1000}\\
4\cdot 20 &\equiv& 17 \pmod{21}\\
\end{array}
\right\}$$

 

.
 Jul 31, 2015
 #1
avatar+26367 
+5
Best Answer

$$\small{
\begin{array}{llrl}
$(A)$ &\text{Find a number }$0\leq x< &14 &$that solves the congruence $9x \equiv 9 \pmod{14}$.\\
$(B)$&\text{Find a number }$0\leq x< &5 &$that solves the congruence $3x \equiv 4 \pmod{5}$.\\
$(C)$&\text{Find a number }$0\leq x< &1000&$that solves the congruence $999x \equiv 998 \pmod{1000}$.\\
$(D)$&\text{Find a number }$0\leq x< &21 &$that solves the congruence $4x \equiv 17 \pmod{21}$.
\end{array}
}$$

 

$$\left\{
\begin{array}{l}
$(A)$\\
$(B)$\\
$(C)$\\
$(D)$
\end{array}
\right\}=
\left\{
\begin{array}{lcr}
x&=&1\\
x&=&3\\
x&=&2\\
x&=&20
\end{array}
\right\}, \small{\text{ because }
\left\{
\begin{array}{lcr}
9\cdot 1 &\equiv& 9 \pmod{14}\\
3\cdot 3 &\equiv& 4 \pmod{5}\\
999\cdot 2 &\equiv& 998 \pmod{1000}\\
4\cdot 20 &\equiv& 17 \pmod{21}\\
\end{array}
\right\}$$

 

heureka Jul 31, 2015

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