[Toán 9] Bài tập hay

T

transformers123

Gọi $A$ là số đã cho, theo đề bài, ta có:

$A\ \equiv\ 3\ \pmod{4}$ nên $A=4a+3\ (a \in N^*)$

Tương tự ta có: $A=17b+9=19c+13\ (b \in N^*, c \in N^*)$

Xét $\begin{cases}A+25=4a+3+25=4(a+7)\ \vdots\ 4\\A+25=17b+9+25=17(b+2)\vdots\
17\\A+25=19c+13+25= 19(c+2)\ \vdots\ 19\end{cases}$

Mà $(4;17;19)=1$ nên $A+25\ \vdots\ 4.17.19$

$\Longrightarrow A+25\ \vdots 1292$

$\Longrightarrow A\ \equiv\ 1267\ \pmod{1292}$
 
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S

soccan

$A \equiv 13\ (mod\ 19) \longrightarrow A=19k+13\\
19k+13 \equiv 9\ (mod\ 17) \longrightarrow 17k+2k+4 \equiv 0\ (mod\ 17) \longrightarrow k+2 \equiv 0\ (mod\ 17)\\
\longrightarrow k=17t-2$
thế vào được $A=19(17t-2)+13=323t-25$
$323t-25 \equiv 3\ (mod\ 4) \longrightarrow 4|t$, đặt $t=4n$
$\longrightarrow A=1292n-25 \equiv -25 \equiv 1267\ (mod\ 1292)\\
(k,n,t \in \mathbb Z^+)$
 
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