Can anyone explain how to solve this kind of questions?

Question

Here's the question:

Find an appropriate solution for x.

≡ 2 ( 3)

≡ 2 ( 5)

≡ 5 ( 7)

≡ 7 ( 8)

I saw some examples like this question...but I still don't know what to do. Thanks in advance.

Answer 1

As $$ x \equiv 2 \ (\mod 3), $$ so there exists an integer $n_1$ such that $$ x = 3n_1 + 2. \tag{1} $$ Then
$$ x \equiv 2 \ ( \mod 5) $$ becomes $$ 3n_1 + 2 \equiv 2 \ ( \mod 5), $$ which implies $$ 3n_1 \equiv 0 \ ( \mod 5), $$ and since $\gcd(3, 5) = 1$, we obtain $$ n_1 \equiv 0 \ ( \mod 5), $$ and thus there exists an integer $n_2$ such that $$ n_1 = 5n_2, $$ and then (1) gives $$ x = 3n_1 = 3 \left( 5 n_2 \right) = 15 n_2. \tag{2} $$ Then $$ x \equiv 5 \ (\mod 7) $$ becomes $$ 15 n_2 \equiv 5 \ ( \mod 7), $$ and since $\gcd( 5, 7) = 1$, we obtain $$ 3n_2 \equiv 1 \ ( \mod 7), $$ and since $$ 1 \equiv 15 \ ( \mod 7), $$ we obtain $$ 3n_2 \equiv 15 \ ( \mod 7), $$ and since $\gcd( 3, 7) = 1$, we have $$ n_2 \equiv 5 \ ( \mod 7), $$ and thus there exists an integer $n_3$ such that $$ n_2 = 7n_3 + 5, $$ and then (2) gives $$ x = 15 n_2 = 15 \left( 7 n_3 + 5 \right) = 105 n_3 + 75. \tag{3} $$ Then $$ x \equiv 7 \ (\mod 8) $$ becomes $$ 105 n_3 + 75 \equiv 7 \ (\mod 8), $$ and since $7 \equiv 15 \ ( \mod 8)$, we have $$ 105 n_3 + 75 \equiv 15 \ ( \mod 8), $$ which implies $$ 105 n_3 \equiv -60 \ ( \mod 8), $$ and since $\gcd( 15, 8) = 1$, we obtain $$ 7n_3 \equiv -4 \ (\mod 8), $$ and since $-4 \equiv 4 \ ( \mod 8)$, we obtain $$ 7 n_3 \equiv 4 \ ( \mod 8), $$ and since $4 \equiv 28\ ( \mod 8)$, we can write $$ 7 n_3 \equiv 28 \ ( \mod 8), $$ and since $\gcd(7, 8) = 1$, therefore we get $$ n_3 \equiv 4 \ ( \mod 8), $$ and so there exists an integer $n_4$ such that $$ n_3 = 8n_4 + 4, $$ and then (3) gives $$ x = 105 n_3 + 75 = 105 \left( 8n_4 + 4 \right) + 75 = 840n_4 + 495, $$ which implies that $$ x \equiv 495 \ ( \mod 840). $$

Hope this helps.